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文章目录
前言
学会二叉树的层序遍历,可以一口气打完以下十题
一、力扣144. 二叉树的前序遍历’
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<>();
if(root == null){
return res;
}
Deque<TreeNode> deq = new LinkedList<>();
TreeNode p = root;
while(!deq.isEmpty() || p != null){
if(p != null){
res.add(p.val);
deq.offerLast(p);
p = p.left;
}else{
p = deq.pollLast();
p = p.right;
}
}
return res;
}
}
二、力扣94. 二叉树的中序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<>();
Deque<TreeNode> deq = new LinkedList<>();
if(root == null){
return res;
}
TreeNode p = root;
while(!deq.isEmpty() || p != null){
if(p != null){
deq.offerLast(p);
p = p.left;
}else{
p = deq.pollLast();
res.add(p.val);
p = p.right;
}
}
return res;
}
}
三、力扣145. 二叉树的后序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new LinkedList<>();
if(root == null){
return res;
}
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
TreeNode cur = deq.pollLast();
res.add(cur.val);
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
Collections.reverse(res);
return res;
}
}
四、力扣102. 二叉树的层序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;3
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Deque<TreeNode> deq = new LinkedList<>();
List<List<Integer>> res = new ArrayList<>();
if(root == null){
return res;
}
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
list.add(cur.val);
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
res.add(list);
}
return res;
}
}
五、力扣107. 二叉树的层序遍历 II
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
Deque<TreeNode> deq = new LinkedList<>();
if(root == null){
return res;
}
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size;i ++){
TreeNode cur = deq.pollFirst();
list.add(cur.val);
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
res.add(list);
}
Collections.reverse(res);
return res;
}
}
六、力扣199. 二叉树的右视图
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
if(root == null){
return res;
}
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
for(int i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
if(i == size-1){
res.add(cur.val);
}
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
}
return res;
}
}
七、力扣637. 二叉树的层平均值
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Double> averageOfLevels(TreeNode root) {
List<Double> res = new ArrayList<>();
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
double size = deq.size();
double sum = 0;
for(double i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
sum += cur.val;
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
res.add(sum/size);
}
return res;
}
}
八、力扣429. N 叉树的层序遍历
/*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
};
*/
class Solution {
public List<List<Integer>> levelOrder(Node root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null){
return res;
}
Deque<Node> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i ++){
Node cur = deq.pollFirst();
list.add(cur.val);
for(Node node : cur.children){
if(node != null){
deq.offerLast(node);
}
}
}
res.add(list);
}
return res;
}
}
九、力扣515. 在每个树行中找最大值
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> largestValues(TreeNode root) {
List<Integer> res = new ArrayList<>();
if(root == null){
return res;
}
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
int max = Integer.MIN_VALUE;
int size = deq.size();
for(int i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
max = max > cur.val ? max : cur.val;
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
res.add(max);
}
return res;
}
}
十、力扣116. 填充每个节点的下一个右侧节点指针
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
Deque<Node> deq = new LinkedList<>();
if(root == null){
return root;
}
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
Node pre = deq.peekFirst();
for(int i = 0; i <size; i ++){
Node cur = deq.pollFirst();
if(i > 0){
pre.next = cur;
pre = cur;
}
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
}
return root;
}
}
十一、力扣117. 填充每个节点的下一个右侧节点指针 II
/*
// Definition for a Node.
class Node {
public int val;
public Node left;
public Node right;
public Node next;
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, Node _left, Node _right, Node _next) {
val = _val;
left = _left;
right = _right;
next = _next;
}
};
*/
class Solution {
public Node connect(Node root) {
if(root == null){
return root;
}
Deque<Node> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
int size = deq.size();
Node pre = deq.peekFirst();
for(int i = 0; i < size; i ++){
Node cur = deq.pollFirst();
if(i > 0){
pre.next = cur;
pre = cur;
}
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
}
return root;
}
}
十二、力扣04. 二叉树的最大深度
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if(root == null){
return 0;
}
int res = 0;
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
res ++;
int size = deq.size();
for(int i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
}
return res;
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
return fun(root);
}
public int fun(TreeNode root){
if(root == null){
return 0;
}
int l = fun(root.left);
int r = fun(root.right);
return l > r ? l + 1 : r + 1;
}
}
十三、力扣111. 二叉树的最小深度
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if(root == null){
return 0;
}
int res = 0;
Deque<TreeNode> deq = new LinkedList<>();
deq.offerLast(root);
while(!deq.isEmpty()){
res ++;
int size = deq.size();
for(int i = 0; i < size; i ++){
TreeNode cur = deq.pollFirst();
if(cur.left == null && cur.right == null){
return res;
}
if(cur.left != null){
deq.offerLast(cur.left);
}
if(cur.right != null){
deq.offerLast(cur.right);
}
}
}
return res;
}
}