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Symmetric Tree 深度优先搜索
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 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}".
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Tree Depth-first Search 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
/** * 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 dfscheck(TreeNode *leftNode,TreeNode *rightNode){ if(leftNode==NULL &&rightNode==NULL) return true; if(leftNode==NULL || rightNode==NULL) return false; return leftNode->val==rightNode->val && dfscheck(leftNode->left,rightNode->right) && dfscheck(leftNode->right,rightNode->left); } bool isSymmetric(TreeNode *root) { if(root==NULL) return true; return dfscheck(root->left,root->right); }};
Symmetric Tree 深度优先搜索
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