SymmetricTree

题目描述[原题描述][https://www.acwing.com/problem/content/description/38/]

请实现一个函数,用来判断一棵二叉树是不是对称的。

如果一棵二叉树和它的镜像一样,那么它是对称的。

样例

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如下图所示二叉树[1,2,2,3,4,4,3,null,null,null,null,null,null,null,null]为对称二叉树:
1
/ \
2 2
/ \ / \
3 4 4 3

如下图所示二叉树[1,2,2,null,4,4,3,null,null,null,null,null,null]不是对称二叉树:
1
/ \
2 2
\ / \
4 4 3

算法描述

两边同时遍历,判断两个子树是否为镜像,一次递归,直到遍历到树的子节点,返回结果

C++代码

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/**
* 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:
bool isSymmetric(TreeNode* root) {
if(root==NULL)return true;
return dfs(root->left,root->right);
}

bool dfs(TreeNode* p1,TreeNode* p2){
if(!p1&&!p2)return true;
if(!p1||!p2)return false;
if(p1->val!=p2->val)return false;
return dfs(p1->left,p2->right)&&dfs(p1->right,p2->left);
}
};

Java代码

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

boolean dfs(TreeNode p,TreeNode q){
if(p==null&&q==null)return true;
if(p==null||q==null)return false;
if(p.val!=q.val)return false;
return dfs(p.left,q.right)&&dfs(p.right,q.left);
}
}