98. 验证二叉搜索树

目录

• 题目描述
• 思路
• 代码实现

题目描述

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}".

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

思路

如果一颗二叉树是一颗二叉搜索树，那么对这棵树进行中序遍历，得到的是一个递增排序的访问序列,中序遍历，递归方法,思路类似于剑指offer中将二叉搜索树BST变为递增排序的双向链表的思路.

代码实现

class Solution {
public:
    bool isValidBST(TreeNode *root) {

    	if(root == nullptr)
    		return true;
    	TreeNode * pre = nullptr;       
        return isValidBSTCore(root,pre);        
    }

    bool isValidBSTCore(TreeNode *root,TreeNode * & pre){

    	if(root->left)
    	{
    		if(!isValidBSTCore(root->left,pre))
    			return false;
    	}

    	if(pre!=nullptr && root->val <= pre->val)
    		return false;       
        pre = root;
        if(root->right)
        	return isValidBSTCore(root->right,pre);
        return true;

    	}
};