解题思路:
对二叉树进行反转,如果根节点为空,则返回
创建临时节点,将右节点赋值给临时节点,将左节点赋值给右节点,再将临时节点赋值给左节点。
递归交换左右子树。
返回根节点。
LeetCode官方题解:官方题解
/**
* 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) {
if(root==null)
{
return root;
}
//交换左右子树
TreeNode temp=root.right;
root.right=root.left;
root.left=temp;
//递归交换左右子树
invertTree(root.left);
invertTree(root.right);
return root;
}
}