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.
Solution: Recursion. 1. Add lower & upper bound. O(n)
2. Inorder traversal with one additional parameter (value of predecessor).
O(n)
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 int val = INT_MIN; 14 return isValidBSTRe(root, val); 15 } 16 17 bool isValidBSTRe(TreeNode* root, int& val) 18 { 19 if(!root) return true; 20 21 if(root->left && !isValidBSTRe(root->left, val)) 22 return false; 23 if(root->val <= val) 24 return false; 25 val = root->val; 26 if(root->right && !isValidBSTRe(root->right, val)) 27 return false; 28 return true; 29 } 30 };
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原文：http://www.cnblogs.com/zhengjiankang/p/3649673.html