Leetcode: Balanced Binary Tree

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.
Below solution uses two recursion, which is not recommended:
find the max depth of left and right node separately, and compare to see their difference. Actually recursion is repeated here, which
can be improved.

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}

int left = maxDepth(root.left);
int right = maxDepth(root.right);
if (Math.abs(left - right) > 1) {
return false;
} else {
return isBalanced(root.left) && isBalanced(root.right);
}

}

private int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}

return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
A preferable method is: when calculating for the max depth, exam whether the difference of depth of left subtree and right subtree is larger than 1 at the same time. Once it is, return -1 as a false answer.

/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
return maxDepth(root) != -1;
}

private int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}

int left = maxDepth(root.left);
int right = maxDepth(root.right);
if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
return -1;
}

return Math.max(left, right) + 1;
}
}