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面试题 27. 二叉树的镜像

题目描述

请完成一个函数，输入一个二叉树，该函数输出它的镜像。

例如输入：

     4
   /   \
  2     7
 / \   / \
1   3 6   9

镜像输出：

     4
   /   \
  7     2
 / \   / \
9   6 3   1

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

限制：

• 0 <= 节点个数 <= 1000

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def mirrorTree(self, root: TreeNode) -> TreeNode:
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.mirrorTree(root.left)
        self.mirrorTree(root.right)
        return root

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if (root == null) return null;
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        mirrorTree(root.left);
        mirrorTree(root.right);
        return root;
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var mirrorTree = function (root) {
  if (!root) return null;
  [root.left, root.right] = [root.right, root.left];
  mirrorTree(root.left);
  mirrorTree(root.right);
  return root;
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func mirrorTree(root *TreeNode) *TreeNode {
    if root == nil {
        return root
    }
    root.Left, root.Right = root.Right, root.Left
    mirrorTree(root.Left)
    mirrorTree(root.Right)
    return root
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* mirrorTree(TreeNode* root) {
        // 后续遍历
        if (nullptr == root) {
            return nullptr;
        }

        mirrorTree(root->left);
        mirrorTree(root->right);
        std::swap(root->left, root->right);

        return root;
    }
};

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