0098. Validate Binary Search Tree¶

Problem¶

Leetcode	98. Validate Binary Search Tree
Difficulty	Medium
Tags	Tree, DFS, BST

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as: - Left subtree contains only nodes with keys less than the node's key. - Right subtree contains only nodes with keys greater than the node's key. - Both subtrees must also be BSTs.

Examples¶

Input: root = [2,1,3]   → true
Input: root = [5,1,4,null,null,3,6]   → false

Approach 1: Recursive with Bounds¶

Pass lo, hi bounds; node value must be strictly between.

Approach 2: Inorder Traversal¶

A BST inorder produces strictly increasing values. Track previous node.

Complexity¶

• Time: O(n); Space: O(h)

Implementation¶

Go¶

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func isValidBST(root *TreeNode) bool {
    var dfs func(n *TreeNode, lo, hi int64) bool
    dfs = func(n *TreeNode, lo, hi int64) bool {
        if n == nil { return true }
        v := int64(n.Val)
        if v <= lo || v >= hi { return false }
        return dfs(n.Left, lo, v) && dfs(n.Right, v, hi)
    }
    return dfs(root, -1<<62, 1<<62)
}

Java¶

class Solution {
    public boolean isValidBST(TreeNode root) {
        return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    private boolean dfs(TreeNode n, long lo, long hi) {
        if (n == null) return true;
        if (n.val <= lo || n.val >= hi) return false;
        return dfs(n.left, lo, n.val) && dfs(n.right, n.val, hi);
    }
}

Python¶

class Solution:
    def isValidBST(self, root):
        def dfs(n, lo, hi):
            if not n: return True
            if n.val <= lo or n.val >= hi: return False
            return dfs(n.left, lo, n.val) and dfs(n.right, n.val, hi)
        return dfs(root, float('-inf'), float('inf'))

Visual Animation¶

animation.html