110. Balanced Binary Tree

Problem

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 depth of the two subtrees of every node never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

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 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;
        }
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isBalanced = function(root) {
  return helper(root, 0) >= 0;
};

var helper = function (root, depth) {
  if (!root) return depth;
  var left = helper(root.left, depth + 1);
  var right = helper(root.right, depth + 1);
  if (left === -1 || right === -1 || Math.abs(left - right) > 1) return -1;
  return Math.max(left, right);
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).