symmetric-tree

最新推荐文章于 2022-11-21 09:51:22 发布

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本文介绍了一种检查二叉树是否为中心对称的方法，通过递归算法判断树的左子树和右子树是否互为镜像。文章提供了一个具体的实现案例，并探讨了递归和迭代两种解决方案。

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题目描述

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3

But the following is not:
1
/ \
2 2
\ \
3 3

Note:
Bonus points if you could solve it both recursively and iteratively.
confused what”{1,#,2,3}”means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.
Here’s an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

/**
 * Definition for binary tree
 * 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  recurjudge(root->left,root->right);        
    }
    bool recurjudge(TreeNode* left,TreeNode* right)
        {
        if(left==NULL&&right==NULL)
            return true;
        if(left==NULL||right==NULL)
            return false;
        if(left->val!=right->val)
            return false;
        return recurjudge(left->left,right->right)&recurjudge(left->right,right->left);
    }

};