226. 翻转二叉树
https://leetcode.cn/problems/invert-binary-tree/
难度简单1361
给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。
示例 1：
输入：root = [4,2,7,1,3,6,9] 输出：[4,7,2,9,6,3,1]
示例 2：
输入：root = [2,1,3] 输出：[2,3,1]
示例 3：
输入：root = [] 输出：[]
提示：
- 树中节点数目范围在
[0, 100]内 -100 <= Node.val <= 100
通过次数501,076提交次数631,866
/**
* 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 TreeNode invertTree(TreeNode root) {
//递归翻转
if(root==null) return root;
TreeNode node= root.right;
root.right = root.left;
root.left = node;
invertTree(root.left);
invertTree(root.right);
return root;
}
}/**
* 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 TreeNode invertTree(TreeNode root) {
//广度优先遍历 + 右->左
if(root==null) return root;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty())
{
int size = queue.size();
while(size>0)
{
TreeNode node = queue.poll();
TreeNode temp = node.right;
node.right = node.left;
node.left = temp;
if(node.right!=null) queue.offer(node.right);
if(node.left!=null) queue.offer(node.left);
size--;
}
}
return root;
}
}
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