Description

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
/ \
2 2
/ \ / \
3 4 4 3

But the following [1,2,2,null,3,null,3] is not:

    1
/ \
2 2
\ \
3 3

Note:
Bonus points if you could solve it both recursively and iteratively.

分析

题目的意思是:判断二叉树是否是对称的。

  • 二叉树基本是递归,不仅要判断对称点的值是否相等,也要判断一个节点存在,对称节点不存在的情况。递归的终止条件为两个节点都为空。

代码

/**
* 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){
return true;
}
return isEqual(root->left,root->right);
}

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

参考文献

​[编程题]symmetric-tree​