[LeetCode从零单刷]Symmetric Tree

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题目:

 

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
解答:

最开始非常脑残地希望补全二叉树,然后利用1,2,4,8个节点之间的关系来比较。结果越来越复杂。

还是那句话:碰上树,能递归就递归,不能递归就栈、队列。这次的递归要点:某节点的左子树的左子树 = 右子树的右子树,左子树的右子树 = 右子树的左子树。

 

/** * 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 compare(TreeNode* left, TreeNode* right) { if (!left && !right) return true; if((!left && right) || (left && !right) || (left->val != right->val)) return false; return compare(left->left, right->right) && compare(left->right, right->left); } bool isSymmetric(TreeNode* root) { if (root == NULL || (!root->left && !root->right)) return true; else return compare(root->left, root->right); }};

 

 



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