判断二叉树是否是平衡二叉树

/**

* Definition of TreeNode:

 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

class ResultType{
    
    public boolean isBalanced;
    public int maxDepth;
    
    public ResultType(boolean isBalanced, int maxDepth)
    {
        this.isBalanced = isBalanced;
        this.maxDepth = maxDepth;
    }
}

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    public boolean isBalanced(TreeNode root) {
        // write your code here
        
        return helper(root).isBalanced;
    }
    
    private ResultType helper(TreeNode root)
    {
        if(root == null)
        {
            return new ResultType(true, 0);
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        
        // subtree not balance
        if(!left.isBalanced || !right.isBalanced)
            return new ResultType(false, -1);
        
        // root not balance
        if(Math.abs(left.maxDepth - right.maxDepth) > 1)
            return new ResultType(false, -1);
            
        
        return new ResultType(true, Math.max(left.maxDepth, right.maxDepth)+1);
    }
    
    
}

　　

93. 平衡二叉树

给定一个二叉树,确定它是高度平衡的。对于这个问题,一棵高度平衡的二叉树的定义是：一棵二叉树中每个节点的两个子树的深度相差不会超过1。 

样例

样例  1:
	输入: tree = {1,2,3}
	输出: true
	
	样例解释:
	如下，是一个平衡的二叉树。
		  1  
		 / \                
		2  3

	
样例  2:
	输入: tree = {3,9,20,#,#,15,7}
	输出: true
	
	样例解释:
	如下，是一个平衡的二叉树。
		  3  

https://www.lintcode.cn/problem/balanced-binary-tree/description