offer-55-2

输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。

示例 1:

给定二叉树 [3,9,20,null,null,15,7]

  3
 / \
9  20
  /  \
 15   7

返回 true 。

示例 2:

给定二叉树 [1,2,2,3,3,null,null,4,4]

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

返回 false 。

限制：

• 0 <= 树的结点个数 <= 10000

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }
        return getBalancedDepth(root) != -1;
    }


    private int getBalancedDepth(TreeNode node) {
        if (node == null) {
            return 0;
        }

        int left = getBalancedDepth(node.left);
        if (left == -1) {
            return -1;
        }
        int right = getBalancedDepth(node.right);
        if (right == -1) {
            return -1;
        }

        int gap = left - right;

        if (gap < -1 || gap > 1) {
            return -1;
        }
        return left > right ? left + 1 : right + 1;
    }
}
//leetcode submit region end(Prohibit modification and deletion)