LeetCode之“树”：Validate Binary Search Tree

　　题目链接

　　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：