二叉树的镜像

问题描述

操作给定的二叉树，将其变换为源二叉树的镜像。

来源：牛客网
链接：https://www.nowcoder.com/practice/564f4c26aa584921bc75623e48ca3011?tpId=13&tqId=11171&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

C++代码实现

思路：递归交换左右子树

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    void Mirror(TreeNode *pRoot) {
        if (pRoot)
        {
            auto node = pRoot->left;
            pRoot->left = pRoot->right;
            pRoot->right = node;
            Mirror(pRoot->left);
            Mirror(pRoot->right);
        }
    }
};

python2代码实现

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回镜像树的根节点
    def Mirror(self, root):
        # write code here
        if root is not None:
            node = root.left
            root.left = root.right
            root.right = node
            self.Mirror(root.left)
            self.Mirror(root.right)

java代码实现

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    public void Mirror(TreeNode root) {
        if (root != null)
        {
            TreeNode node = root.left;
            root.left = root.right;
            root.right = node;
            Mirror(root.left);
            Mirror(root.right);
        }
    }
}