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判断一个二叉树是不是自己的镜像

时间:2014-09-04 23:36:10      阅读:308      评论:0      收藏:0      [点我收藏+]

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bool helper(TreeNode *pA, TreeNode *pB) {
        if (!pA && !pB) return true;
        if (!pA || !pB) return false;  // only one has node in a tree and b tree
        if (pA->val != pB->val) return false;
        return helper(pA->left, pB->.right) && helper(pA->right, pB->left); // a goes in in-order traversal, b goes right first then left.
    }
    
public:
    bool isSymmetric(TreeNode *root) {
        if (!root)
            return true;
        
        return helper(root->left, root->right);
    }

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.

判断一个二叉树是不是自己的镜像

标签:style   blog   color   ar   strong   for   div   sp   log   

原文地址:http://www.cnblogs.com/zhhwgis/p/3956965.html

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