import java.util.ArrayList;
import java.util.Deque;
import java.util.LinkedList;
import java.util.List;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
// 构造函数
TreeNode() {}
TreeNode(int val)
{ this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
// 递归写法
//class Solution {
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> res = new ArrayList<>();
// if(root == null)return res;
// preorder(root,res);
// return res;
// }
//
// public void preorder(TreeNode root, List<Integer> res){
// // 终止条件
// if(root == null)return;
// // 将左子树的值放入列表中,将根节点的值放入列表中,将右子树的值放入列表中
// preorder(root.left,res);
// res.add(root.val);
// preorder(root.right,res);
//
// }
//}
class Solution1{
public List<Integer> inorderTraversal(TreeNode root){
List<Integer> res = new ArrayList<>();
if(root == null)return res;
Deque<TreeNode> stack = new LinkedList<>();
while(!stack.isEmpty() || root != null){
while(root != null){
stack.push(root);
root = root.left;
}
root = stack.pop();
res.add(root.val);
root = root.right;
}
return res;
}
}
class test{
public static void main(String[] args) {
TreeNode node1 = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
TreeNode node6 = new TreeNode(6);
TreeNode node7 = new TreeNode(7);
node1.left = node2;node1.right = node3;
node2.left = node4;node2.right = node5;
node3.left = node6;node3.right = node7;
List<Integer> res = new ArrayList<>();
res = new Solution1().inorderTraversal(node1);
for (int i = 0; i < res.size(); i++) {
System.out.println(res.get(i));
}
}
}
2、写代码思路
因为学习了前序遍历的迭代代码,因此,我把中序遍历的迭代方法当作一个考试题来做,借助各种技巧来做;
两个while;
其中一个while是栈不为空且节点不为空;
另一个while是节点不为空;
还是要建立一个数组和一个栈;
栈中要存放的是根节点;
数组存放的中序遍历的值;
问题是什么时候把root存进栈和什么时候把root.val放进数组中;
对比一下,其实就是数组存放中序遍历的值变了;
3、中序遍历和前序遍历的代码对比
(1)中序遍历
class Solution1{
public List<Integer> inorderTraversal(TreeNode root){
List<Integer> res = new ArrayList<>();
if(root == null)return res;
Deque<TreeNode> stack = new LinkedList<>();
while(!stack.isEmpty() || root != null){
while(root != null){
stack.push(root);
root = root.left;
}
root = stack.pop();
res.add(root.val);
root = root.right;
}
return res;
}
}
(2)前序遍历
class Solution1{
public List<Integer> preorderTraversal(TreeNode root){
List<Integer> list = new ArrayList<>();
if(root == null)return list;
Deque<TreeNode> stack = new LinkedList<>();
while(!stack.isEmpty() || root != null){
// 注意这里是双重循环,而不是单循环
while(root != null){
stack.push(root);
list.add(root.val);
root = root.left;
}
root = stack.pop();
root = root.right;
}
return list;
}
}