98. Validate Binary Search Tree - Medium

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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

 

M1: recursion

给每个节点定义一个区间，如果节点的值落在区间外，返回false，否则继续递归左右节点，区间上下限相应调整

注意！！min, max如果用int会溢出，要用long

time: O(n), space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    public boolean isValid(TreeNode node, long min, long max) {
        if(node == null) {
            return true;
        }
        
        if(node.val <= min || node.val >= max) {
            return false;
        }
        return isValid(node.left, min, node.val) && isValid(node.right, node.val, max);
    }
}

 

M2: iteration using inorder traversal

用一个指针时刻记录root移动前的状态，因为BST的inorder traversal是升序，如果prev.val >= cur.val，返回false；否则继续遍历

time = O(n), space = O(n) worst case

class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) {
            return true;
        }
        LinkedList<TreeNode> stack = new LinkedList<>();
        TreeNode prev = null, cur = root;
        while(!stack.isEmpty() || cur != null) {
            while(cur != null) {
                stack.offerFirst(cur);
                cur = cur.left;
            }
            cur = stack.pollFirst();
            if(prev != null && prev.val >= cur.val) {
                return false;
            }
            prev = cur;
            cur = cur.right;
        }
        return true;
    }
}