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Java-based LeetCode algorithm problem solutions, regularly updated
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98\. Validate Binary Search Tree
Medium
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:**
![](https://assets.leetcode.com/uploads/2020/12/01/tree1.jpg)
**Input:** root = [2,1,3]
**Output:** true
**Example 2:**
![](https://assets.leetcode.com/uploads/2020/12/01/tree2.jpg)
**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]
.
* -231 <= Node.val <= 231 - 1
To solve the "Validate Binary Search Tree" problem in Java with the Solution class, follow these steps:
1. Define a method `isValidBST` in the `Solution` class that takes the root of a binary tree as input and returns true if the tree is a valid binary search tree (BST), and false otherwise.
2. Implement a recursive approach to validate if the given binary tree is a valid BST:
- Define a helper method `isValidBSTHelper` that takes the root node, a lower bound, and an upper bound as input parameters.
- In the `isValidBSTHelper` method, recursively traverse the binary tree nodes.
- At each node, check if its value is within the specified bounds (lower bound and upper bound) for a valid BST.
- If the node's value violates the BST property, return false.
- Otherwise, recursively validate the left and right subtrees by updating the bounds accordingly.
- If both the left and right subtrees are valid BSTs, return true.
3. Call the `isValidBSTHelper` method with the root node and appropriate initial bounds to start the validation process.
Here's the implementation of the `isValidBST` method in Java:
```java
class Solution {
public boolean isValidBST(TreeNode root) {
return isValidBSTHelper(root, null, null);
}
private boolean isValidBSTHelper(TreeNode node, Integer lower, Integer upper) {
if (node == null) {
return true;
}
int val = node.val;
if ((lower != null && val <= lower) || (upper != null && val >= upper)) {
return false;
}
return isValidBSTHelper(node.left, lower, val) && isValidBSTHelper(node.right, val, upper);
}
}
```
This implementation recursively validates whether the given binary tree is a valid BST in O(n) time complexity, where n is the number of nodes in the tree.
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