翻转一棵二叉树。
示例:
输入:
4 / \ 2 7 / \ / \ 1 3 6 9
输出:
4 / \ 7 2 / \ / \ 9 6 3 1
备注:
这个问题是受到 Max Howell 的 原问题 启发的 :
谷歌:我们90%的工程师使用您编写的软件(Homebrew),但是您却无法在面试时在白板上写出翻转二叉树这道题,这太糟糕了。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if root is None:
return None
root.left, root.right = root.right, root.left
self.invertTree(root.left)
self.invertTree(root.right)
return root
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) return null;
TreeNode t = root.left;
root.left = root.right;
root.right = t;
invertTree(root.left);
invertTree(root.right);
return root;
}
}
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
var invertTree = function (root) {
if (!root) return null;
[root.left, root.right] = [root.right, root.left];
invertTree(root.left);
invertTree(root.right);
return root;
};
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
if (root == nil) {
return nil
}
root.Left, root.Right = root.Right, root.Left
invertTree(root.Left)
invertTree(root.Right)
return root
}