【转】"《剑指offer》刷题笔记(面试思路):二叉树的镜像"



题目描述

操作给定的二叉树,将其变换为源二叉树的镜像。

二叉树的镜像定义:

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源二叉树 
8
/ \
6 10
/ \ / \
5 7 9 11
镜像二叉树
8
/ \
10 6
/ \ / \
11 9 7 5

解题思路

还是BFS和DFS的套路,要么递归实现要么利用队列进行层序遍历。

C++版代码实现

递归

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/*
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){
swap(pRoot->left, pRoot->right);
Mirror(pRoot->left);
Mirror(pRoot->right);
}
}
};

层序遍历

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/*
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) return;
TreeNode* node;
queue<TreeNode*> que;
que.push(pRoot);
while(!que.empty()){
node = que.front();
que.pop();
swap(node->left, node->right);
if(node->left) que.push(node->left);
if(node->right) que.push(node->right);
}
}
};

Python版代码实现

递归

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# -*- 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 != None:
root.left,root.right = root.right,root.left
self.Mirror(root.left)
self.Mirror(root.right)

层序遍历

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# -*- 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 not root: return
a=[root]
while a:
b=[]
for node in a:
node.left,node.right = node.right,node.left
if node.left:
b.append(node.left)
if node.right:
b.append(node.right)
a=b

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完的汪(∪。∪)。。。zzz