Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than the node's key.
- Both the left and right subtrees must also be binary search trees.
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}"
.
中序遍历一遍就可以了
/** * 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 isValidBSTsub(TreeNode* root,TreeNode** pre){ if(root==NULL)return true; if(!isValidBSTsub(root->left,pre))return false; if((*pre)!=NULL){ if((*pre)->val>=root->val)return false; } *pre=root; if(!isValidBSTsub(root->right,pre))return false; return true; } bool isValidBST(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function TreeNode* pre=NULL; return isValidBSTsub(root,&pre); } };