Validate Binary Search Tree

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.

思路：可用递归解决，我觉得唯一的难点在于要保证每次递归前，左子树的值均小于根植，右子树的值均大于根植。

（左右子树NULL节点均为满足判断条件的点）

/**
 * 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 greatThanVal(int v,TreeNode *root)
    {
        if(NULL == root)
        return true;
        if(root->val > v)
        return greatThanVal(v,root->left)&&greatThanVal(v,root->right);
        return false;
    }
    bool lessThanVal(int v,TreeNode *root)
    {
        if(NULL == root)
        return true;
        if(root->val < v)
        return lessThanVal(v,root->left)&&lessThanVal(v,root->right);
        return false;
    }
    bool isValidBST(TreeNode* root) {
        if(NULL == root)
        return true;
        if(NULL == root->left && NULL == root->right)
        return true;
        if(NULL == root->left)
        {
              if(greatThanVal(root->val,root->right))
              return isValidBST(root->right);
              else return false;
        }
       
        if(NULL == root->right)
        {
            if(lessThanVal(root->val,root->left))
            return isValidBST(root->left);
            else
            return false;
        }
        if(greatThanVal(root->val,root->right)&&lessThanVal(root->val,root->left))
            return isValidBST(root->left)&&isValidBST(root->right);
        else
            return false;
    }
};