# 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.

# Java 实现

``````
package com.coderli.leetcode.algorithms.easy;

/**
* Given a binary tree, determine if it is height-balanced.
* <p>
* 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.
*
* @author OneCoder 2017-11-23 22:05
*/
public class BalancedBinaryTree {

public static void main(String[] args) {
BalancedBinaryTree balancedBinaryTree = new BalancedBinaryTree();
TreeNode tree = balancedBinaryTree.new TreeNode(1);
TreeNode subNodeLeft = balancedBinaryTree.new TreeNode(2);
TreeNode subNodeRight = balancedBinaryTree.new TreeNode(2);
subNodeLeft.left = balancedBinaryTree.new TreeNode(3);
subNodeLeft.right = balancedBinaryTree.new TreeNode(4);
subNodeRight.left = balancedBinaryTree.new TreeNode(4);
subNodeRight.right = balancedBinaryTree.new TreeNode(3);
tree.left = subNodeLeft;
tree.right = subNodeRight;
System.out.println(balancedBinaryTree.isBalanced(tree));
}

public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
int leftSubTreeHeight = treeHeight(root.left);
int rightSubTreeHeight = treeHeight(root.right);
int differ = leftSubTreeHeight - rightSubTreeHeight;
if ( differ > 1 || differ < -1) {
return false;
}
return isBalanced(root.left) && isBalanced(root.right);
}

private int treeHeight(TreeNode tree) {
if (tree == null) {
return 0;
}
int leftSubHeight = treeHeight(tree.left);
int rightSubHeight = treeHeight(tree.right);
return leftSubHeight >= rightSubHeight ? leftSubHeight + 1: rightSubHeight + 1;
}

public class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}
}

``````

Thanks a lot.