g0101_0200.s0110_balanced_binary_tree.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g0101_0200.s0110_balanced_binary_tree;

// #Easy #Depth_First_Search #Tree #Binary_Tree #Programming_Skills_II_Day_2 #Level_2_Day_6_Tree
// #Udemy_Tree_Stack_Queue #2022_06_22_Time_1_ms_(98.82%)_Space_44.8_MB_(19.20%)

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
/**
 * 110 - Balanced Binary Tree\.
 *
 * Easy
 *
 * Given a binary tree, determine if it is height-balanced.
 *
 * For this problem, a height-balanced binary tree is defined as:
 *
 * > a binary tree in which the left and right subtrees of _every_ node differ in height by no more than 1.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/06/balance_1.jpg)
 *
 * **Input:** root = [3,9,20,null,null,15,7]
 *
 * **Output:** true 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/06/balance_2.jpg)
 *
 * **Input:** root = [1,2,2,3,3,null,null,4,4]
 *
 * **Output:** false 
 *
 * **Example 3:**
 *
 * **Input:** root = []
 *
 * **Output:** true 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range `[0, 5000]`.
 * *   -10⁴ <= Node.val <= 10⁴
**/
public class Solution {
    public boolean isBalanced(TreeNode root) {
        // Empty Tree is balanced
        if (root == null) {
            return true;
        }
        // Get max height of subtree child
        // Get max height of subtree child
        // compare height difference (cannot be more than 1)
        int leftHeight = 0;
        int rightHeight = 0;
        if (root.left != null) {
            leftHeight = getMaxHeight(root.left);
        }
        if (root.right != null) {
            rightHeight = getMaxHeight(root.right);
        }
        int heightDiff = Math.abs(leftHeight - rightHeight);
        // if passes height check
        //  - Check if left subtree is balanced and if the right subtree is balanced
        //  - If one of both are imbalanced, then the tree is imbalanced
        return heightDiff <= 1 && isBalanced(root.left) && isBalanced(root.right);
    }

    private int getMaxHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = 0;
        int rightHeight = 0;
        // Left
        if (root.left != null) {
            leftHeight = getMaxHeight(root.left);
        }
        // Right
        if (root.right != null) {
            rightHeight = getMaxHeight(root.right);
        }
        if (leftHeight > rightHeight) {
            return 1 + leftHeight;
        } else {
            return 1 + rightHeight;
        }
    }
}