Java二叉树算法总结-优快云博客

package Algorithms.tree;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Stack;
/**
* REFS:
* http://blog.youkuaiyun.com/fightforyourdream/article/details/16843303 面试大总结之二：Java搞定面试中的二叉树题目
* http://blog.youkuaiyun.com/luckyxiaoqiang/article/details/7518888 轻松搞定面试中的二叉树题目
* http://www.cnblogs.com/Jax/archive/2009/12/28/1633691.html 算法大全（3）二叉树
*
* 1. 求二叉树中的节点个数: getNodeNumRec（递归），getNodeNum（迭代）
* 2. 求二叉树的深度: getDepthRec（递归），getDepth
* 3. 前序遍历，中序遍历，后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* 4.分层遍历二叉树（按层次从上往下，从左往右）: levelTraversal, levelTraversalRec（递归解法）
* 5. 将二叉查找树变为有序的双向链表: convertBST2DLLRec, convertBST2DLL
* 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* 7. 求二叉树中叶子节点的个数：getNodeNumLeafRec, getNodeNumLeaf
* 8. 判断两棵二叉树是否相同的树：isSameRec, isSame
* 9. 判断二叉树是不是平衡二叉树：isAVLRec
* 10. 求二叉树的镜像（破坏和不破坏原来的树两种情况）：
* mirrorRec, mirrorCopyRec
* mirror, mirrorCopy
* 10.1 判断两个树是否互相镜像：isMirrorRec isMirror
* 11. 求二叉树中两个节点的最低公共祖先节点：
* LAC 求解最小公共祖先, 使用list来存储path.
* LCABstRec 递归求解BST树.
* LCARec 递归算法 .
* 12. 求二叉树中节点的最大距离：getMaxDistanceRec
* 13. 由前序遍历序列和中序遍历序列重建二叉树：rebuildBinaryTreeRec
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTree, isCompleteBinaryTreeRec
* 15. 找出二叉树中最长连续子串(即全部往左的连续节点，或是全部往右的连续节点）findLongest
*/
public class TreeDemo {
/*
1
/ \
2 3
/ \ \
4 5 6
*/
public static void main(String[] args) {
TreeNode r1 = new TreeNode(1);
TreeNode r2 = new TreeNode(2);
TreeNode r3 = new TreeNode(3);
TreeNode r4 = new TreeNode(4);
TreeNode r5 = new TreeNode(5);
TreeNode r6 = new TreeNode(6);
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
/*
1
/ \
2 3
/ \ \
4 5 6
*/
// TreeNode r1 = new TreeNode(10);
// TreeNode r2 = new TreeNode(6);
// TreeNode r3 = new TreeNode(14);
// TreeNode r4 = new TreeNode(4);
// TreeNode r5 = new TreeNode(8);
// TreeNode r6 = new TreeNode(16);
TreeNode r7 = new TreeNode(0);
r1.left = r2;
r1.right = r3;
r2.left = r4;
r2.right = r5;
r3.right = r6;
r4.left = r7;
TreeNode t1 = new TreeNode(10);
TreeNode t2 = new TreeNode(6);
TreeNode t3 = new TreeNode(14);
TreeNode t4 = new TreeNode(4);
TreeNode t5 = new TreeNode(8);
TreeNode t6 = new TreeNode(16);
TreeNode t7 = new TreeNode(0);
TreeNode t8 = new TreeNode(0);
TreeNode t9 = new TreeNode(0);
TreeNode t10 = new TreeNode(0);
TreeNode t11 = new TreeNode(0);
t1.left = t2;
t1.right = t3;
t2.left = t4;
t2.right = t5;
t3.left = t6;
t3.right = t7;
t4.left = t8;
//t4.right = t9;
t5.right = t9;
// test distance
// t5.right = t8;
// t8.right = t9;
// t9.right = t10;
// t10.right = t11;
/*
10
/ \
6 14
/ \ \
4 8 16
/
0
*/
// System.out.println(LCABstRec(t1, t2, t4).val);
// System.out.println(LCABstRec(t1, t2, t6).val);
// System.out.println(LCABstRec(t1, t4, t6).val);
// System.out.println(LCABstRec(t1, t4, t7).val);
// System.out.println(LCABstRec(t1, t3, t6).val);
//
// System.out.println(LCA(t1, t2, t4).val);
// System.out.println(LCA(t1, t2, t6).val);
// System.out.println(LCA(t1, t4, t6).val);
// System.out.println(LCA(t1, t4, t7).val);
// System.out.println(LCA(t1, t3, t6).val);
// System.out.println(LCA(t1, t6, t6).val);
//System.out.println(getMaxDistanceRec(t1));
//System.out.println(isSame(r1, t1));
// System.out.println(isAVLRec(r1));
//
// preorderTraversalRec(r1);
// //mirrorRec(r1);
// //TreeNode r1Mirror = mirror(r1);
//
// TreeNode r1MirrorCopy = mirrorCopy(r1);
// System.out.println();
// //preorderTraversalRec(r1Mirror);
// preorderTraversalRec(r1MirrorCopy);
//
// System.out.println();
//
// System.out.println(isMirrorRec(r1, r1MirrorCopy));
// System.out.println(isMirror(r1, r1MirrorCopy));
//System.out.println(getNodeNumKthLevelRec(r1, 5));
//System.out.println(getNodeNumLeaf(r1));
// System.out.println(getNodeNumRec(null));
// System.out.println(getNodeNum(r1));
//System.out.println(getDepthRec(null));
// System.out.println(getDepth(r1));
//
// preorderTraversalRec(r1);
// System.out.println();
// preorderTraversal(r1);
// System.out.println();
// inorderTraversalRec(r1);
//
// System.out.println();
// inorderTraversal(r1);
// postorderTraversalRec(r1);
// System.out.println();
// postorderTraversal(r1);
// System.out.println();
// levelTraversal(r1);
//
// System.out.println();
// levelTraversalRec(r1);
// TreeNode ret = convertBST2DLLRec(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// TreeNode ret2 = convertBST2DLL(r1);
// while (ret2.right != null) {
// ret2 = ret2.right;
// }
//
// while (ret2 != null) {
// System.out.print(ret2.val + " ");
// ret2 = ret2.left;
// }
//
// TreeNode ret = convertBST2DLL(r1);
// while (ret != null) {
// System.out.print(ret.val + " ");
// ret = ret.right;
// }
// System.out.println();
// System.out.println(findLongest(r1));
// System.out.println();
// System.out.println(findLongest2(r1));
// test the rebuildBinaryTreeRec.
//test_rebuildBinaryTreeRec();
System.out.println(isCompleteBinaryTreeRec(t1));
System.out.println(isCompleteBinaryTree(t1));
}
public static void test_rebuildBinaryTreeRec() {
ArrayList<Integer> list1 = new ArrayList<Integer>();
list1.add(1);
list1.add(2);
list1.add(4);
list1.add(5);
list1.add(3);
list1.add(6);
list1.add(7);
list1.add(8);
ArrayList<Integer> list2 = new ArrayList<Integer>();
list2.add(4);
list2.add(2);
list2.add(5);
list2.add(1);
list2.add(3);
list2.add(7);
list2.add(6);
list2.add(8);
TreeNode root = rebuildBinaryTreeRec(list1, list2);
preorderTraversalRec(root);
System.out.println();
postorderTraversalRec(root);
}
private static class TreeNode{
int val;
TreeNode left;
TreeNode right;
public TreeNode(int val){
this.val = val;
left = null;
right = null;
}
}
/*
* null返回0，然后把左右子树的size加上即可。
* */
public static int getNodeNumRec(TreeNode root) {
if (root == null) {
return 0;
}
return getNodeNumRec(root.left) + getNodeNumRec(root.right) + 1;
}
/**
* 求二叉树中的节点个数迭代解法O(n)：基本思想同LevelOrderTraversal，
* 即用一个Queue，在Java里面可以用LinkedList来模拟
*/
public static int getNodeNum(TreeNode root) {
if (root == null) {
return 0;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
int cnt = 0;
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
cnt++;
}
return cnt;
}
public static int getDepthRec(TreeNode root) {
if (root == null) {
return -1;
}
return Math.max(getDepthRec(root.left), getDepthRec(root.right)) + 1;
}
/*
* 可以用 level LevelOrderTraversal 来实现，我们用一个dummyNode来分隔不同的层，这样即可计算出实际的depth.
* 1
/ \
2 3
/ \ \
4 5 6
*
* 在队列中如此排列： 1, dummy, 2, 3, dummy, 4, 5, 5, dummy
*
*/
public static int getDepth(TreeNode root) {
if (root == null) {
return 0;
}
TreeNode dummy = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummy);
int depth = -1;
while (!q.isEmpty()) {
TreeNode curr = q.poll();
if (curr == dummy) {
depth++;
if (!q.isEmpty()) { // 使用DummyNode来区分不同的层，如果下一层不是为空，则应该在尾部加DummyNode.
q.offer(dummy);
}
}
if (curr.left != null) {
q.offer(curr.left);
}
if (curr.right != null) {
q.offer(curr.right);
}
}
return depth;
}
/*
* 3. 前序遍历，中序遍历，后序遍历: preorderTraversalRec, preorderTraversal, inorderTraversalRec, postorderTraversalRec
* (https://en.wikipedia.org/wiki/Tree_traversal#Pre-order_2)
* */
public static void preorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
System.out.print(root.val + " ");
preorderTraversalRec(root.left);
preorderTraversalRec(root.right);
}
/*
* 前序遍历，Iteration 算法. 把根节点存在stack中。
* */
public static void preorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode node = s.pop();
System.out.print(node.val + " ");
if (node.right != null) { //
s.push(node.right);
}
// 我们需要先压入右节点，再压入左节点，这样就可以先弹出左节点。
if (node.left != null) {
s.push(node.left);
}
}
}
/*
* 中序遍历
* */
public static void inorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
inorderTraversalRec(root.left);
System.out.print(root.val + " ");
inorderTraversalRec(root.right);
}
/**
* 中序遍历迭代解法，用栈先把根节点的所有左孩子都添加到栈内，
* 然后输出栈顶元素，再处理栈顶元素的右子树
* http://www.youtube.com/watch?v=50v1sJkjxoc
*
* 还有一种方法能不用递归和栈，基于线索二叉树的方法，较麻烦以后补上
* http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/
*/
public static void inorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while(true) {
// 把当前节点的左节点都push到栈中.
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// 因为此时已经没有左孩子了，所以输出栈顶元素
cur = s.pop();
System.out.print(cur.val + " ");
// 准备处理右子树
cur = cur.right;
}
}
// 后序遍历
/*
* 1
/ \
2 3
/ \ \
4 5 6
if put into the stack directly, then it should be:
1, 2, 4, 5, 3, 6 in the stack.
when pop, it should be: 6, 3, 5, 4, 2, 1
if I
* */
public static void postorderTraversalRec(TreeNode root) {
if (root == null) {
return;
}
postorderTraversalRec(root.left);
postorderTraversalRec(root.right);
System.out.print(root.val + " ");
}
/**
* 后序遍历迭代解法
* http://www.youtube.com/watch?v=hv-mJUs5mvU
* http://blog.youkuaiyun.com/tang_jin2015/article/details/8545457
* 从左到右的后序与从右到左的前序的逆序是一样的，所以就简单喽！哈哈
* 用另外一个栈进行翻转即可喽
*/
public static void postorderTraversal(TreeNode root) {
if (root == null) {
return;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> out = new Stack<TreeNode>();
s.push(root);
while(!s.isEmpty()) {
TreeNode cur = s.pop();
out.push(cur);
if (cur.left != null) {
s.push(cur.left);
}
if (cur.right != null) {
s.push(cur.right);
}
}
while(!out.isEmpty()) {
System.out.print(out.pop().val + " ");
}
}
/*
* 分层遍历二叉树（按层次从上往下，从左往右）迭代
* 其实就是广度优先搜索，使用队列实现。队列初始化，将根节点压入队列。当队列不为空，进行如下操作：弹出一个节点
* ，访问，若左子节点或右子节点不为空，将其压入队列
* */
public static void levelTraversal(TreeNode root) {
if (root == null) {
return;
}
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
TreeNode cur = q.poll();
System.out.print(cur.val + " ");
if (cur.left != null) {
q.offer(cur.left);
}
if (cur.right != null) {
q.offer(cur.right);
}
}
}
public static void levelTraversalRec(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
levelTraversalVisit(root, 0, ret);
System.out.println(ret);
}
/**
* 分层遍历二叉树（递归）
* 很少有人会用递归去做level traversal
* 基本思想是用一个大的ArrayList，里面包含了每一层的ArrayList。
* 大的ArrayList的size和level有关系
*
* http://discuss.leetcode.com/questions/49/binary-tree-level-order-traversal#answer-container-2543
*/
public static void levelTraversalVisit(TreeNode root, int level, ArrayList<ArrayList<Integer>> ret) {
if (root == null) {
return;
}
// 如果ArrayList的层数不够用，则新添加一层
// when size = 3, level: 0, 1, 2
if (level >= ret.size()) {
ret.add(new ArrayList<Integer>());
}
// visit 当前节点
ret.get(level).add(root.val);
// 将左子树，右子树添加到对应的层。
levelTraversalVisit(root.left, level + 1, ret);
levelTraversalVisit(root.right, level + 1, ret);
}
/*
* 题目要求：将二叉查找树转换成排序的双向链表，不能创建新节点，只调整指针。
查找树的结点定义如下：
既然是树，其定义本身就是递归的，自然用递归算法处理就很容易。将根结点的左子树和右子树转换为有序的双向链表，
然后根节点的left指针指向左子树结果的最后一个结点，同时左子树最后一个结点的right指针指向根节点；
根节点的right指针指向右子树结果的第一个结点，
同时右子树第一个结点的left指针指向根节点。
* */
public static TreeNode convertBST2DLLRec(TreeNode root) {
return convertBST2DLLRecHelp(root)[0];
}
/*
* ret[0] 代表左指针
* ret[1] 代表右指针
* */
public static TreeNode[] convertBST2DLLRecHelp(TreeNode root) {
TreeNode[] ret = new TreeNode[2];
ret[0] = null;
ret[1] = null;
if (root == null) {
return ret;
}
if (root.left != null) {
TreeNode left[] = convertBST2DLLRecHelp(root.left);
left[1].right = root; // 将左子树的尾节点连接到根
root.left = left[1];
ret[0] = left[0];
} else {
ret[0] = root; // 左节点返回root.
}
if (root.right != null) {
TreeNode right[] = convertBST2DLLRecHelp(root.right);
right[0].left = root; // 将右子树的头节点连接到根
root.right = right[0];
ret[1] = right[1];
} else {
ret[1] = root; // 右节点返回root.
}
return ret;
}
/**
* 将二叉查找树变为有序的双向链表迭代解法
* 类似inOrder traversal的做法
*/
public static TreeNode convertBST2DLL(TreeNode root) {
while (root == null) {
return null;
}
TreeNode pre = null;
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
TreeNode head = null; // 链表头
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
// if stack is empty, just break;
if (s.isEmpty()) {
break;
}
cur = s.pop();
if (head == null) {
head = cur;
}
// link pre and cur.
cur.left = pre;
if (pre != null) {
pre.right = cur;
}
// 左节点已经处理完了，处理右节点
cur = cur.right;
pre = cur;
}
return root;
}
/*
* * 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevel(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
int level = 0;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
TreeNode dummy = new TreeNode(0);
int cnt = 0; // record the size of the level.
q.offer(dummy);
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node == dummy) {
level++;
if (level == k) {
return cnt;
}
cnt = 0; // reset the cnt;
if (q.isEmpty()) {
break;
}
q.offer(dummy);
continue;
}
cnt++;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
return 0;
}
/*
* * 6. 求二叉树第K层的节点个数：getNodeNumKthLevelRec, getNodeNumKthLevel
* */
public static int getNodeNumKthLevelRec(TreeNode root, int k) {
if (root == null || k <= 0) {
return 0;
}
if (k == 1) {
return 1;
}
// 将左子树及右子树在K层的节点个数相加.
return getNodeNumKthLevelRec(root.left, k - 1) + getNodeNumKthLevelRec(root.right, k - 1);
}
/*
* 7. getNodeNumLeafRec 把左子树和右子树的叶子节点加在一起即可
* */
public static int getNodeNumLeafRec(TreeNode root) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
return 1;
}
return getNodeNumLeafRec(root.left) + getNodeNumLeafRec(root.right);
}
/* 7. getNodeNumLeaf
* 随便使用一种遍历方法都可以，比如，中序遍历。
* inorderTraversal，判断是不是叶子节点。
* */
public static int getNodeNumLeaf(TreeNode root) {
if (root == null) {
return 0;
}
int cnt = 0;
// we can use inorderTraversal travesal to do it.
Stack<TreeNode> s = new Stack<TreeNode>();
TreeNode cur = root;
while (true) {
while (cur != null) {
s.push(cur);
cur = cur.left;
}
if (s.isEmpty()) {
break;
}
// all the left child has been put into the stack, let's deal with the
// current node.
cur = s.pop();
if (cur.left == null && cur.right == null) {
cnt++;
}
cur = cur.right;
}
return cnt;
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 递归解法：
* （1）如果两棵二叉树都为空，返回真
* （2）如果两棵二叉树一棵为空，另一棵不为空，返回假
* （3）如果两棵二叉树都不为空，如果对应的左子树和右子树都同构返回真，其他返回假
* */
public static boolean isSameRec(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
// 1. the value of the root should be the same;
// 2. the left tree should be the same.
// 3. the right tree should be the same.
return r1.val == r2.val &&
isSameRec(r1.left, r2.left) && isSameRec(r1.right, r2.right);
}
/*
* 8. 判断两棵二叉树是否相同的树。
* 迭代解法
* 我们直接用中序遍历来比较就好啦
* */
public static boolean isSame(TreeNode r1, TreeNode r2) {
// both are null.
if (r1 == null && r2 == null) {
return true;
}
// one is null.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
TreeNode cur1 = r1;
TreeNode cur2 = r2;
while (true) {
while (cur1 != null && cur2 != null) {
s1.push(cur1);
s2.push(cur2);
cur1 = cur1.left;
cur2 = cur2.left;
}
if (cur1 != null || cur2 != null) {
return false;
}
if (s1.isEmpty() && s2.isEmpty()) {
break;
}
cur1 = s1.pop();
cur2 = s2.pop();
if (cur1.val != cur2.val) {
return false;
}
cur1 = cur1.right;
cur2 = cur2.right;
}
return true;
}
/*
*
* 9. 判断二叉树是不是平衡二叉树：isAVLRec
* 1. 左子树，右子树的高度差不能超过1
* 2. 左子树，右子树都是平衡二叉树。
*
*/
public static boolean isAVLRec(TreeNode root) {
if (root == null) {
return true;
}
// 左子树，右子树都必须是平衡二叉树。
if (!isAVLRec(root.left) || !isAVLRec(root.right)) {
return false;
}
int dif = Math.abs(getDepthRec(root.left) - getDepthRec(root.right));
if (dif > 1) {
return false;
}
return true;
}
/**
* 10. 求二叉树的镜像递归解法：
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且交换它们
TreeNode tmp = root.right;
root.right = mirrorRec(root.left);
root.left = mirrorRec(tmp);
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法：
*
* (1) 破坏原来的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* */
public static TreeNode mirror(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
s.push(root);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
// 交换当前节点的左右节点
TreeNode tmp = cur.left;
cur.left = cur.right;
cur.right = tmp;
// traversal 左节点，右节点。
if (cur.right != null) {
s.push(cur.right);
}
if (cur.left != null) {
s.push(cur.left);
}
}
return root;
}
/**
* 10. 求二叉树的镜像 Iterator解法：
*
* (2) 创建一个新的树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
*
* 应该可以使用任何一种Traversal 方法。
* 我们现在可以试看看使用最简单的前序遍历。
* 前序遍历我们可以立刻把新建好的左右节点创建出来，比较方便
* */
public static TreeNode mirrorCopy(TreeNode root) {
if (root == null) {
return null;
}
Stack<TreeNode> s = new Stack<TreeNode>();
Stack<TreeNode> sCopy = new Stack<TreeNode>();
s.push(root);
TreeNode rootCopy = new TreeNode(root.val);
sCopy.push(rootCopy);
while (!s.isEmpty()) {
TreeNode cur = s.pop();
TreeNode curCopy = sCopy.pop();
// traversal 左节点，右节点。
if (cur.right != null) {
// copy 在这里做比较好，因为我们可以容易地找到它的父节点
TreeNode leftCopy = new TreeNode(cur.right.val);
curCopy.left = leftCopy;
s.push(cur.right);
sCopy.push(curCopy.left);
}
if (cur.left != null) {
// copy 在这里做比较好，因为我们可以容易地找到它的父节点
TreeNode rightCopy = new TreeNode(cur.left.val);
curCopy.right = rightCopy;
s.push(cur.left);
sCopy.push(curCopy.right);
}
}
return rootCopy;
}
/**
* 10. 求二叉树的镜像递归解法：
*
* (1) 不破坏原来的树，新建一个树
*
* 1 1
* / \
* 2 -----> 2
* \ /
* 3 3
* */
public static TreeNode mirrorCopyRec(TreeNode root) {
if (root == null) {
return null;
}
// 先把左右子树分别镜像,并且把它们连接到新建的root节点。
TreeNode rootCopy = new TreeNode(root.val);
rootCopy.left = mirrorCopyRec(root.right);
rootCopy.right = mirrorCopyRec(root.left);
return rootCopy;
}
/*
* 10.1. 判断两个树是否互相镜像
* (1) 根必须同时为空，或是同时不为空
*
* 如果根不为空:
* (1).根的值一样
* (2).r1的左树是r2的右树的镜像
* (3).r1的右树是r2的左树的镜像
* */
public static boolean isMirrorRec(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空，则返回false.
if (r1 == null || r2 == null) {
return false;
}
// If both are not null, they should be:
// 1. have same value for root.
// 2. R1's left tree is the mirror of R2's right tree;
// 3. R2's right tree is the mirror of R1's left tree;
return r1.val == r2.val
&& isMirrorRec(r1.left, r2.right)
&& isMirrorRec(r1.right, r2.left);
}
/*
* 10.1. 判断两个树是否互相镜像 Iterator 做法
* (1) 根必须同时为空，或是同时不为空
*
* 如果根不为空:
* traversal 整个树，判断它们是不是镜像，每次都按照反向来traversal
* (1). 当前节点的值相等
* (2). 当前节点的左右节点要镜像，
* 无论是左节点，还是右节点，对应另外一棵树的镜像位置，可以同时为空，或是同时不为空，但是不可以一个为空，一个不为空。
* */
public static boolean isMirror(TreeNode r1, TreeNode r2){
// 如果2个树都是空树
if (r1 == null && r2 == null) {
return true;
}
// 如果其中一个为空，则返回false.
if (r1 == null || r2 == null) {
return false;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(r1);
s2.push(r2);
while (!s1.isEmpty() && !s2.isEmpty()) {
TreeNode cur1 = s1.pop();
TreeNode cur2 = s2.pop();
// 弹出的节点的值必须相等
if (cur1.val != cur2.val) {
return false;
}
// tree1的左节点，tree2的右节点，可以同时不为空，也可以同时为空，否则返回false.
TreeNode left1 = cur1.left;
TreeNode right1 = cur1.right;
TreeNode left2 = cur2.left;
TreeNode right2 = cur2.right;
if (left1 != null && right2 != null) {
s1.push(left1);
s2.push(right2);
} else if (!(left1 == null && right2 == null)) {
return false;
}
// tree1的左节点，tree2的右节点，可以同时不为空，也可以同时为空，否则返回false.
if (right1 != null && left2 != null) {
s1.push(right1);
s2.push(left2);
} else if (!(right1 == null && left2 == null)) {
return false;
}
}
return true;
}
/*
* 11. 求二叉树中两个节点的最低公共祖先节点：
* Recursion Version:
* LACRec
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LACRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
// if no node is in the node, just recursively find it in LEFT and RIGHT tree.
TreeNode left = LACRec(root.left, node1, node2);
TreeNode right = LACRec(root.right, node1, node2);
if (left == null) { // If didn't found in the left tree, then just return it from right.
return right;
} else if (right == null) { // Or if didn't found in the right tree, then just return it from the left side.
return left;
}
// if both right and right found a node, just return the root as the Common Ancestor.
return root;
}
/*
* 11. 求BST中两个节点的最低公共祖先节点：
* Recursive version:
* LCABst
*
* 1. If found in the left tree, return the Ancestor.
* 2. If found in the right tree, return the Ancestor.
* 3. If Didn't find any of the node, return null.
* 4. If found both in the left and the right tree, return the root.
* */
public static TreeNode LCABstRec(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || node1 == null || node2 == null) {
return null;
}
// If any of the node is the root, just return the root.
if (root == node1 || root == node2) {
return root;
}
int min = Math.min(node1.val, node2.val);
int max = Math.max(node1.val, node2.val);
// if the values are smaller than the root value, just search them in the left tree.
if (root.val > max) {
return LCABstRec(root.left, node1, node2);
} else if (root.val < min) {
// if the values are larger than the root value, just search them in the right tree.
return LCABstRec(root.right, node1, node2);
}
// if root is in the middle, just return the root.
return root;
}
/*
* 解法1. 记录下path,并且比较之:
* LAC
* http://www.geeksforgeeks.org/lowest-common-ancestor-binary-tree-set-1/
* */
public static TreeNode LCA(TreeNode root, TreeNode r1, TreeNode r2) {
// If the nodes have one in the root, just return the root.
if (root == null || r1 == null || r2 == null) {
return null;
}
ArrayList<TreeNode> list1 = new ArrayList<TreeNode>();
ArrayList<TreeNode> list2 = new ArrayList<TreeNode>();
boolean find1 = LCAPath(root, r1, list1);
boolean find2 = LCAPath(root, r2, list2);
// If didn't find any of the node, just return a null.
if (!find1 || !find2) {
return null;
}
// 注意: 使用Iterator 对于linkedlist可以提高性能。
// 所以统一使用Iterator 来进行操作。
Iterator<TreeNode> iter1 = list1.iterator();
Iterator<TreeNode> iter2 = list2.iterator();
TreeNode last = null;
while (iter1.hasNext() && iter2.hasNext()) {
TreeNode tmp1 = iter1.next();
TreeNode tmp2 = iter2.next();
if (tmp1 != tmp2) {
return last;
}
last = tmp1;
}
// If never find any node which is different, means Node 1 and Node 2 are the same one.
// so just return the last one.
return last;
}
public static boolean LCAPath(TreeNode root, TreeNode node, ArrayList<TreeNode> path) {
// if didn't find, we should return a empty path.
if (root == null || node == null) {
return false;
}
// First add the root node.
path.add(root);
// if the node is in the left side.
if (root != node
&& !LCAPath(root.left, node, path)
&& !LCAPath(root.right, node, path)
) {
// Didn't find the node. should remove the node added before.
path.remove(root);
return false;
}
// found
return true;
}
/*
* * 12. 求二叉树中节点的最大距离：getMaxDistanceRec
*
* 首先我们来定义这个距离：
* 距离定义为：两个节点间边的数目.
* 如：
* 1
* / \
* 2 3
* \
* 4
* 这里最大距离定义为2，4的距离，为3.
* 求二叉树中节点的最大距离即二叉树中相距最远的两个节点之间的距离。 (distance / diameter)
* 递归解法：
* 返回值设计：
* 返回1. 深度， 2. 当前树的最长距离
* (1) 计算左子树的深度，右子树深度，左子树独立的链条长度，右子树独立的链条长度
* (2) 最大长度为三者之最：
* a. 通过根节点的链，为左右深度+2
* b. 左子树独立链
* c. 右子树独立链。
*
* (3)递归初始条件：
* 当root == null, depth = -1.maxDistance = -1;
*
*/
public static int getMaxDistanceRec(TreeNode root) {
return getMaxDistanceRecHelp(root).maxDistance;
}
public static Result getMaxDistanceRecHelp(TreeNode root) {
Result ret = new Result(-1, -1);
if (root == null) {
return ret;
}
Result left = getMaxDistanceRecHelp(root.left);
Result right = getMaxDistanceRecHelp(root.right);
// 深度应加1， the depth from the subtree to the root.
ret.depth = Math.max(left.depth, right.depth) + 1;
// 左子树，右子树与根的距离都要加1，所以通过根节点的路径为两边深度+2
int crossLen = left.depth + right.depth + 2;
// 求出cross根的路径，及左右子树的独立路径，这三者路径的最大值。
ret.maxDistance = Math.max(left.maxDistance, right.maxDistance);
ret.maxDistance = Math.max(ret.maxDistance, crossLen);
return ret;
}
private static class Result {
int depth;
int maxDistance;
public Result(int depth, int maxDistance) {
this.depth = depth;
this.maxDistance = maxDistance;
}
}
/*
* 13. 由前序遍历序列和中序遍历序列重建二叉树：rebuildBinaryTreeRec
* We assume that there is no duplicate in the trees.
* For example:
* 1
* / \
* 2 3
* /\ \
* 4 5 6
* /\
* 7 8
*
* PreOrder should be: 1 2 4 5 3 6 7 8
* 根左子树右子树
* InOrder should be: 4 2 5 1 3 7 6 8
* 左子树根右子树
* */
public static TreeNode rebuildBinaryTreeRec(List<Integer> preOrder, List<Integer> inOrder) {
if (preOrder == null || inOrder == null) {
return null;
}
// If the traversal is empty, just return a NULL.
if (preOrder.size() == 0 || inOrder.size() == 0) {
return null;
}
// we can get the root from the preOrder.
// Because the first one is the root.
// So we just create the root node here.
TreeNode root = new TreeNode(preOrder.get(0));
List<Integer> preOrderLeft;
List<Integer> preOrderRight;
List<Integer> inOrderLeft;
List<Integer> inOrderRight;
// 获得在 inOrder中，根的位置
int rootInIndex = inOrder.indexOf(preOrder.get(0));
preOrderLeft = preOrder.subList(1, rootInIndex + 1);
preOrderRight = preOrder.subList(rootInIndex + 1, preOrder.size());
// 得到inOrder左边的左子树
inOrderLeft = inOrder.subList(0, rootInIndex);
inOrderRight = inOrder.subList(rootInIndex + 1, inOrder.size());
// 通过 Rec 来调用生成左右子树。
root.left = rebuildBinaryTreeRec(preOrderLeft, inOrderLeft);
root.right = rebuildBinaryTreeRec(preOrderRight, inOrderRight);
return root;
}
/*
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTree, isCompleteBinaryTreeRec
* 进行level traversal, 一旦遇到一个节点的左节点为空，后面的节点的子节点都必须为空。而且不应该有下一行，其实就是队列中所有的
* 元素都不应该再有子元素。
* */
public static boolean isCompleteBinaryTree(TreeNode root) {
if (root == null) {
return false;
}
TreeNode dummyNode = new TreeNode(0);
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
q.offer(dummyNode);
// if this is true, no node should have any child.
boolean noChild = false;
while (!q.isEmpty()) {
TreeNode cur = q.poll();
if (cur == dummyNode) {
if (!q.isEmpty()) {
q.offer(dummyNode);
}
// Dummy node不需要处理。
continue;
}
if (cur.left != null) {
// 如果标记被设置，则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.left);
} else {
// 一旦某元素没有左节点或是右节点，则之后所有的元素都不应有子元素。
// 并且该元素不可以有右节点.
noChild = true;
}
if (cur.right != null) {
// 如果标记被设置，则Queue中任何元素不应再有子元素。
if (noChild) {
return false;
}
q.offer(cur.right);
} else {
// 一旦某元素没有左节点或是右节点，则之后所有的元素都不应有子元素。
noChild = true;
}
}
return true;
}
/*
* 14. 判断二叉树是不是完全二叉树：isCompleteBinaryTreeRec
*
*
* 我们可以分解为：
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件：
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件：
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static boolean isCompleteBinaryTreeRec(TreeNode root) {
return isCompleteBinaryTreeRecHelp(root).isCompleteBT;
}
private static class ReturnBinaryTree {
boolean isCompleteBT;
boolean isPerfectBT;
int height;
ReturnBinaryTree(boolean isCompleteBT, boolean isPerfectBT, int height) {
this.isCompleteBT = isCompleteBT;
this.isPerfectBT = isPerfectBT;
this.height = height;
}
}
/*
* 我们可以分解为：
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
*
* Base 条件：
* (1) root = null: 为perfect & complete BinaryTree, Height -1;
*
* 而 Perfect Binary Tree的条件：
* 左右子树均为Perfect Binary Tree,并且Height 相同。
* */
public static ReturnBinaryTree isCompleteBinaryTreeRecHelp(TreeNode root) {
ReturnBinaryTree ret = new ReturnBinaryTree(true, true, -1);
if (root == null) {
return ret;
}
ReturnBinaryTree left = isCompleteBinaryTreeRecHelp(root.left);
ReturnBinaryTree right = isCompleteBinaryTreeRecHelp(root.right);
// 树的高度为左树高度，右树高度的最大值+1
ret.height = 1 + Math.max(left.height, right.height);
// set the isPerfectBT
ret.isPerfectBT = false;
if (left.isPerfectBT && right.isPerfectBT && left.height == right.height) {
ret.isPerfectBT = true;
}
// set the isCompleteBT.
/*
* CompleteBinary Tree 的条件是：
* 1. 左右子树均为Perfect binary tree, 并且两者Height相同(其实就是本树是perfect tree)
* 2. 左子树为CompleteBinaryTree, 右子树为Perfect binary tree，并且两者Height差1
* 3. 左子树为Perfect Binary Tree,右子树为CompleteBinaryTree, 并且Height 相同
* */
ret.isCompleteBT = ret.isPerfectBT
|| (left.isCompleteBT && right.isPerfectBT && left.height == right.height + 1)
|| (left.isPerfectBT && right.isCompleteBT && left.height == right.height);
return ret;
}
/*
* 15. findLongest
* 第一种解法：
* 返回左边最长，右边最长，及左子树最长，右子树最长。
* */
public static int findLongest(TreeNode root) {
if (root == null) {
return -1;
}
TreeNode l = root;
int cntL = 0;
while (l.left != null) {
cntL++;
l = l.left;
}
TreeNode r = root;
int cntR = 0;
while (r.right != null) {
cntR++;
r = r.right;
}
int lmax = findLongest(root.left);
int rmax = findLongest(root.right);
int max = Math.max(lmax, rmax);
max = Math.max(max, cntR);
max = Math.max(max, cntL);
return max;
}
/* 1
* 2 3
* 3 4
* 6 1
* 7
* 9
* 11
* 2
* 14
* */
public static int findLongest2(TreeNode root) {
int [] maxVal = new int[1];
maxVal[0] = -1;
findLongest2Help(root, maxVal);
return maxVal[0];
}
// ret:
// 0: the left side longest,
// 1: the right side longest.
static int maxLen = -1;
static int[] findLongest2Help(TreeNode root, int[] maxVal) {
int[] ret = new int[2];
if (root == null) {
ret[0] = -1;
ret[1] = -1;
return ret;
}
ret[0] = findLongest2Help(root.left, maxVal)[0] + 1;
ret[1] = findLongest2Help(root.right, maxVal)[1] + 1;
//maxLen = Math.max(maxLen, ret[0]);
//maxLen = Math.max(maxLen, ret[1]);
maxVal[0] = Math.max(maxVal[0], ret[0]);
maxVal[0] = Math.max(maxVal[0], ret[1]);
return ret;
}
}