【LeetCode】98. Validate Binary Search Tree

Difficulty:Medium

 More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/validate-binary-search-tree/

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.

Intuition

1.Method 1:

　　* Determine if left subtree and right subtree are valid binary search trees

　　* Compare root.val with maxValue in left subtree and minValue in right subtree;

2.Method 2:

　　* give the maxium range {max, min} for root.val

　　* update the {max, min} for subtree‘s root.

Solution

Method 1

    public boolean isValidBST(TreeNode root) {
        if(root==null )
            return true;
        TreeNode left = root.left, right=root.right;
        if(!isValidBST(left) || !isValidBST(right))
            return false;

        // find the maxNode in left subtree
        if(left!=null)
            while(left.right!=null)
                left=left.right;

        // find the minNode in right subtree
        if(right!=null)
            while(right.left!=null)
                right=right.left;

        return (left==null || left.val<root.val)
            &&(right==null || right.val>root.val);
    }

Method 2

    public boolean isValidBST(TreeNode root) {
        return helper(root, (long)Integer.MIN_VALUE-1, (long)Integer.MAX_VALUE+1);
    }

    private boolean helper(TreeNode node, long min, long max){
        if(node == null)
            return true;
        return (node.val < max && node.val>min)
            && helper(node.left, min, node.val)
            && helper(node.right, node.val, max);
    }

Complexity

Method 1:

Time complexity : O(nlogn)

Space complexity : O(n)

Method 2:

Time complexity : O(n)

Space complexity : O(n)

What I‘ve learned

 More:【目录】LeetCode Java实现

原文地址：https://www.cnblogs.com/yongh/p/11750691.html