问题描述
操作给定的二叉树,将其变换为源二叉树的镜像。左右交换
思路1
先判断树是否为空,树的左右结点是否为空
然后交换左右结点,使用递归一直操作到叶子结点
C++实现
/*
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==NULL){
return;
}
if(pRoot->left==NULL&&pRoot->right==NULL){
return;
}
TreeNode *temp;
temp=pRoot->left;
pRoot->left=pRoot->right;
pRoot->right=temp;
Mirror(pRoot->left);
Mirror(pRoot->right);
}
};
java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){
return;
}
if(root.left==null&&root.right==null){
return;
}
TreeNode temp=root.left;
root.left=root.right;
root.right=temp;
Mirror(root.left);
Mirror(root.right);
}
}
python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==None:
return None
if root.left==None and root.right==None:
return root
root.left,root.right=root.right,root.left
self.Mirror(root.left)
self.Mirror(root.right)
思路二
使用循环
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
void swap(TreeNode *root){
TreeNode *temp=root->left;
root->left=root->right;
root->right=temp;
}
void Mirror(TreeNode *pRoot) {
if(pRoot==NULL){
return;
}
if(pRoot->left==NULL&&pRoot->right==NULL){
return;
}
stack<TreeNode *> s;
s.push(pRoot); //根结点进栈
while(!s.empty()){
TreeNode *root1=s.top();
s.pop();
swap(root1);
if(root1->left){
s.push(root1->left);
}
if(root1->right){
s.push(root1->right);
}
}
}
};