# 【LeetCode】 98. Validate Binary Search Tree

## Description
> 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.

> 給予一個二元樹，判斷它是否為合法的二元搜索樹(BST)。

> 一個BST的定義為以下：
> * 左邊的子樹只會包含比你小的key的節點。
> * 右邊的子樹只會包含比你大的key的節點。
> * 左子樹和右子樹也都是二元搜索樹。

## Example:

Example 1:

    2
   / \
  1   3

Input: [2,1,3]
Output: true


Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

## Solution

* 單純使用遞迴去跑所有節點即可，讓自己去和所有子樹去比大小，如果到`NULL`(空節點)為止都沒有錯誤就回傳`true`。

### Code

/**
 * 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 large(TreeNode* root, TreeNode* n)
    {
        if(!n)
            return true;
        if(root->val <= n->val)
            return false;
        return large(root, n->left) && large(root, n->right);
    }
    
    bool small(TreeNode* root, TreeNode* n)
    {
        if(!n)
            return true;
        if(root->val >= n->val)
            return false;
        return small(root, n->left) && small(root, n->right);
    }
    
    bool isValidBST(TreeNode* root) {
        if(!root)
            return true;
        if((large(root, root->left) && small(root, root->right)) == false)
            return false;
        return isValidBST(root->right) && isValidBST(root->left);
    }
};

###### tags: `LeetCode` `C++`