problem:
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.
thinking:
(1)该题合适的解法是递归法,采用DFS思想
递归,保存左右两个节点,然后判断leftNode->left和rightNode->right,以及leftNode->right和rightNode->left。如此不断递归
(2)还想到另外一种非递归法:层序遍历法,子结点为NULL时代表数值0,从左往右层序遍历得到数组a,从右往左层序遍历得到数组b
如果a==b,则该二叉树是镜像的。但提交时显示:Status:
code:
递归法:
class Solution { public: bool check(TreeNode *leftNode, TreeNode *rightNode) { if (leftNode == NULL && rightNode == NULL) return true; if (leftNode == NULL || rightNode == NULL) return false; return leftNode->val == rightNode->val && check(leftNode->left, rightNode->right) && check(leftNode->right, rightNode->left); } bool isSymmetric(TreeNode *root) { if (root == NULL) return true; return check(root->left, root->right); } };层序遍历法:Status:
class Solution { private: vector<int> res1; vector<int> res2; public: bool isSymmetric(TreeNode *root) { if(root==NULL) return true; res1.clear(); res2.clear(); level_walk_left(root); level_walk_right(root); if(res1==res2) return true; else return false; } protected: void level_walk_left(TreeNode *root) { queue<TreeNode *> _queue; _queue.push(root); while(!_queue.empty()) { TreeNode *tmp=_queue.front(); if(tmp==NULL) res1.push_back(0); else { res1.push_back(tmp->val); _queue.pop(); if(tmp->left!=NULL) _queue.push(tmp->left); else _queue.push(NULL); if(tmp->right!=NULL) _queue.push(tmp->right); else _queue.push(NULL); } } } void level_walk_right(TreeNode *root) { queue<TreeNode *> _queue; _queue.push(root); while(!_queue.empty()) { TreeNode *tmp=_queue.front(); if(tmp==NULL) res2.push_back(0); else { res2.push_back(tmp->val); _queue.pop(); if(tmp->right!=NULL) _queue.push(tmp->right); else _queue.push(NULL); if(tmp->left!=NULL) _queue.push(tmp->left); else _queue.push(NULL); } } } };
leetcode || 101、Symmetric Tree
原文地址:http://blog.csdn.net/hustyangju/article/details/45146775