【LeetCode】98. Validate Binary Search Tree

Difficulty:Medium

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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.

3. Method 3:

　　using iterative inorder traversal. 

 

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);
    }

　　

Method 3

    public boolean isValidBST(TreeNode root) {
        if(root==null)
            return true;
        Stack<TreeNode> stk = new Stack<>();
        TreeNode pre = null;
        while(root!=null || !stk.isEmpty()){
            while(root!=null){
                stk.push(root);
                root=root.left;
            }
            TreeNode cur = stk.peek();
            if(pre!=null && pre.val>=cur.val)
                return false;
            root=stk.pop().right;
            pre = cur;
        }
        return true;
    }

　　

Complexity

Method 1:

Time complexity : O(nlogn)

Space complexity : O(n)

Method 2:

Time complexity : O(n)

Space complexity : O(n)

Method 3:

Time complexity : O(n)

Space complexity : O(logn)  depth of the tree

 

 

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