Problem
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.
Solution (Java)
/**
* 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) {
return solve(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
// we will send a valid range and check whether the root lies in the range
// and update the range for the subtrees
private boolean solve(TreeNode root, long left, long right) {
if (root == null) {
return true;
}
if (root.val <= left || root.val >= right) {
return false;
}
return solve(root.left, left, root.val) && solve(root.right, root.val, right);
}
}
Solution 1 (Javascript)
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isValidBST = function(root) {
return helper(root, Number.MIN_SAFE_INTEGER, Number.MAX_SAFE_INTEGER);
};
var helper = function (root, min, max) {
if (!root) return true;
if (root.val <= min || root.val >= max) return false;
return helper(root.left, min, root.val) && helper(root.right, root.val, max);
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).
Solution 2 (Javascript)
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isValidBST = function(root) {
var prev = null;
var now = root;
var stack = [];
while (now || stack.length) {
while (now) {
stack.push(now);
now = now.left;
}
now = stack.pop();
if (prev && prev.val >= now.val) return false;
prev = now;
now = now.right;
}
return true;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).