二叉树的镜像 🌟🌟🌟简单

课后作业

问题描述

原文链接：剑指 Offer 27. 二叉树的镜像

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：

镜像输出：

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

代码实现

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 || (root.left == null && root.right == null)){
            return root;
        }

        TreeNode left = mirrorTree(root.left);
        TreeNode right = mirrorTree(root.right);

        root.left = right;
        root.right = left;

        return root;
    }
}

Python

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

class Solution(object):
    def mirrorTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root or (not root.left and not root.right):
            return root

        left = self.mirrorTree(root.left)
        right = self.mirrorTree(root.right)

        root.left = right
        root.right = left

        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) {
        if (!root || (!root->left && !root->right)) {
            return root;
        }

        TreeNode* left = mirrorTree(root->left);
        TreeNode* right = mirrorTree(root->right);

        root->left = right;
        root->right = left;

        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 || (root.Left == nil && root.Right == nil) {
        return root
    }

    left := mirrorTree(root.Left)
    right := mirrorTree(root.Right)

    root.Left = right
    root.Right = left

    return root
}

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