leetcode78 Validate Binary Search Tree

问题描述

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

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.

分析

这是一道判断二叉树是否为二叉搜索树的问题。
二叉搜索树的特点是：任一节点的左子树都要小于该节点右子树的值。
于是我们可以采用中序遍历的方法，如果是二叉搜索树，则满足按中序遍历的顺序遍历的每个节点的值为递增序列。所以我们中序遍历，并用一个vector来记录每个节点的值。最后遍历vector看是不是按照递增顺序排列。

/**
 * 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:
    void Inorder(TreeNode* node, vector<int>& v)
    {
        if(node->left) Inorder(node->left,v);
        v.push_back(node->val);
        if(node->right)  Inorder(node->right,v);
        return;
    }
    bool isValidBST(TreeNode* root) {
        if(!root) return true;
        vector<int> v;
        Inorder(root, v);
        for(int i = 0; i<v.size()-1; i++)
        {
            if(v[i]>=v[i+1]) return false;
        }
        return true;
    }
};