LEETCODE ALGORITHM:98. Validate Binary Search Tree

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题目

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:

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    2
   / \
  1   3

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

Example 2:

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    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.

题解

中序遍历并检查是否为严格递增序列

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/**
 * 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 {
private:
    vector<int> m;
    void dfs(TreeNode* root)
    {
        if(root==NULL) return;
        dfs(root->left);
        m.push_back(root->val);
        dfs(root->right);
    }
public:
    bool isValidBST(TreeNode* root) {
        dfs(root);
        for(int i=1;i<m.size();i++)
        {
            if(m[i]<=m[i-1]) return false;
        }
        return true;
    }
};
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//只保存中序遍历的上一个值，当前遍历的值小于等于上一个值返回false，反之返回true，节约时间和空间
/**
 * 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 {
private:
    int tmp,flag=0;
    bool dfs(TreeNode* root)
    {
        if(root==NULL) return true;
        if(!dfs(root->left)) return false;
        if(flag!=0&&root->val<=tmp) return false;
        tmp=root->val;++flag;
        if(!dfs(root->right)) return false;
        return true;
    }
public:
    bool isValidBST(TreeNode* root) {
        return dfs(root);
    }
};