leetcode笔记：Validate Binary Search Tree

Posted 2020-10-03 yfceshi

一. 题目描写叙述

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?

二. 题目分析

这道题的大意是，推断一个二叉查找树是否合法的，这个能够依据二叉查找树的定义来进行推断，即一个内部结点的值要大于左子树的最大值，同一时候要小于右子树的最大值。

依据这个定义。能够递归得进行推断，这样的方法得时间复杂度为O(n)，空间复杂度为O(logn)。这道题要注意的地方就是要记得更新以结点为父节点的树的最大值和最小值。以便递归返回时给上一个调用进行推断。

三. 演示样例代码

#include <iostream>

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution
{
private:
    bool isValidBST(TreeNode* root, int &MinValue, int &MaxValue)
    {
        if (!root)
        {
            return true;
        }


        if (root->left)
        {
            if (root->val <= root->left->val)
            {
                return false;
            }

            int LeftMinValue = 0;
            int LeftMaxValue = 0;
            if (!isValidBST(root->left, LeftMinValue, LeftMaxValue))
            {
                return false;
            }
            else
            {
                MinValue = LeftMinValue;
                if (LeftMaxValue != LeftMinValue)
                {
                    if (root->val <= LeftMaxValue)
                    {
                        return false;
                    }
                }
            }
        }
        else
        {
            MinValue = root->val;
        }


        if (root->right)
        {
            if (root->val >= root->right->val)
            {
                return false;
            }

            int RightMinValue = 0;
            int RightMaxValue = 0;

            if (!isValidBST(root->right, RightMinValue, RightMaxValue))
            {
                return false;
            }
            else
            {
                MaxValue = RightMaxValue;
                if (RightMaxValue != RightMinValue)
                {
                    if (root->val >= RightMinValue)
                    {
                        return false;
                    }
                }
            }
        }
        else
        {
            MaxValue = root->val;
        }

        return true;
    }

public:
    bool isValidBST(TreeNode* root)
    {
        int MinValue = 0;
        int MaxValue = 0;
        bool IsLeaf = true;

        return isValidBST(root, MinValue, MaxValue);
    }
};