LeetCode_110. Balanced Binary Tree

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 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.

package leetcode.easy;

public class BalancedBinaryTree {
	@org.junit.Test
	public void test1() {
		TreeNode tn11 = new TreeNode(3);
		TreeNode tn21 = new TreeNode(9);
		TreeNode tn22 = new TreeNode(20);
		TreeNode tn33 = new TreeNode(15);
		TreeNode tn34 = new TreeNode(7);
		tn11.left = tn21;
		tn11.right = tn22;
		tn21.left = null;
		tn21.right = null;
		tn22.left = tn33;
		tn22.right = tn34;
		tn33.left = null;
		tn33.right = null;
		tn34.left = null;
		tn34.right = null;
		System.out.print(isBalanced(tn11));
	}

	@org.junit.Test
	public void test2() {
		TreeNode tn11 = new TreeNode(1);
		TreeNode tn21 = new TreeNode(2);
		TreeNode tn22 = new TreeNode(2);
		TreeNode tn31 = new TreeNode(3);
		TreeNode tn32 = new TreeNode(3);
		TreeNode tn41 = new TreeNode(4);
		TreeNode tn42 = new TreeNode(4);
		tn11.left = tn21;
		tn11.right = tn22;
		tn21.left = tn31;
		tn21.right = tn32;
		tn22.left = null;
		tn22.right = null;
		tn31.left = tn41;
		tn31.right = tn42;
		tn32.left = null;
		tn32.right = null;
		tn41.left = null;
		tn41.right = null;
		tn42.left = null;
		tn42.right = null;
		System.out.print(isBalanced(tn11));
	}

	public boolean isBalanced(TreeNode root) {
		if (null == root) {
			return true;
		} else {
			return (Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1) && isBalanced(root.left)
					&& isBalanced(root.right);
		}
	}

	private static int maxDepth(TreeNode root) {
		if (null == root) {
			return 0;
		} else {
			return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
		}
	}
}