Problem
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}".
Code
bool isvbst_r(TreeNode *root,int min,int max) { if(!root) return true; if(root->val <= min || root->val >= max) return false; return (isvbst_r(root->right,root->val,max) && isvbst_r(root->left,min,root->val)); } bool isValidBST(TreeNode *root) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. return isvbst_r(root,INT_MIN,INT_MAX); }
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