0098 - Validate Binary Search Tree (Medium)
Problem Link
https://leetcode.com/problems/validate-binary-search-tree/
Problem Statement
Given the root of a binary tree, determine if it is a valid binary search tree (BST).
A valid 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: root = [2,1,3]
Output: true
Example 2:
Input: root = [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.
Constraints
- The number of nodes in the tree is in the range [1, 104].
- -
Approach 1: Preorder Traversal
Time Complexity:
Space Complexity: for recursive stack space
- Python
- Java
- C++
- JavaScript
class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
def validate(root, left, right):
if root is None:
return True
# Validate the condition for each subtree
if root.val <= left or root.val >= right:
return False
# all subtrees left of root should be less than right so pass root.val as right
left = validate(root.left, left, root.val)
# all subtrees right of root should be greater than right so pass root.val as left
right = validate(root.right, root.val, right)
# only if left and right subtrees are valid return true
return left and right
# pass -inf as the left minimum and inf as right maximum initially
return validate(root, -inf, inf)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) return true;
return checkBST(root, null, null);
}
public boolean checkBST(TreeNode root, Integer min, Integer max) {
if (root == null) return true;
// Check the node value with it's parent
// If node is left, then value should be less than or equal to it's parent
// If node is right, then value should be greater than or equal to it's parent
if ((min != null && root.val <= min) || (max != null && root.val >= max)) {
return false;
}
return checkBST(root.left, min, root.val) && checkBST(root.right, root.val, max);
}
}
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool valid(TreeNode* node, long left, long right) {
if (!node) {
return true;
}
if (!(node->val > left && node->val < right)) {
return false;
}
return (valid(node->left, left, node->val) &&
valid(node->right, node->val, right));
}
bool isValidBST(TreeNode* root) {
return valid(root, LONG_MIN, LONG_MAX);
}
};
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isValidBST = function(root) {
function valid(node, left, right) {
if (!node) {
return true;
}
if (!(node.val > left && node.val < right)) {
return false;
}
return (valid(node.left, left, node.val) &&
valid(node.right, node.val, right));
}
return valid(root, parseFloat(-Infinity), parseFloat(Infinity));
};