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.
Analysis:
Tree problem always related to its recursive property. Here to judge whether a binary tree is a binary search tree, the condition is as follows.
root
/
left right
left sub-tree's nodes' value should all less than the root's value
right sub-tree's nodes' valu should all greater than the root's value
And do it recursively.
For the left-subtree, the max value is root->val and for right-subtree, the min value is root->val
Here the construction of initial max and min value is a little tricky by using the bit operation.
Code:

1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isValidBST(TreeNode *root) { 13 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 int max = (1<<31)-1; 16 int min = (1<<31); 17 18 return judge(root, max, min); 19 } 20 21 private: 22 bool judge(TreeNode *n, int max, int min) { 23 if (n == NULL) 24 return true; 25 26 if (n->val < max && n->val > min) 27 return judge(n->left, n->val, min) && judge(n->right, max, n->val); 28 else 29 return false; 30 } 31 };