题目要求：
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}".
用惯性思维（即判断某节点的值与其左右子节点的值得关系）去解决这道题，相对麻烦些（花了比较多的时间，最后还是很难兼顾各种情况）。我们可以用中序遍历的方法将整棵树输出，然后再判断输出序列是否是递增序列就行了。
还有一篇博文的想法特别好，我们要将关注点放在节点本身，然后不断更新它的上下限。
具体程序如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
return isValidBSTSub(root, LONG_MIN, LONG_MAX);
} bool isValidBSTSub(TreeNode *tree, long alpha, long beta)
{
if(!tree)
return true; if(tree->val > alpha && tree->val < beta)
return isValidBSTSub(tree->left, alpha, tree->val) &&
isValidBSTSub(tree->right, tree->val, beta);
else
return false;
}
};
上边程序用到了LONG_MIN、LONG_MAX，主要是为了应付比较极端的测试集。下图是C/C++中个数据类型的最大值宏定义列表：