给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，一棵高度平衡二叉树定义为：

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。

tips :
1.平衡树的左右子树都是平衡树 。
2.空树是平衡树。

下图不是平衡树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        return process(root).isBalanced;
    }

    private Info process(TreeNode root){
        if(root == null){
            return new Info(0,true);
        }

        Info leftInfo = process(root.left);
        Info rightInfo = process(root.right);

        int leftHeight = leftInfo.height;
        int rightHeight = rightInfo.height;
        int height = Math.max(leftHeight,rightHeight) + 1;
        int ans = Math.abs(leftHeight - rightHeight);
        boolean isBalanced = true;
        if(ans > 1){
            isBalanced = false;
        }
        //验证每一个节点的左右孩子是否是平衡树
        if(leftInfo != null){
            isBalanced = isBalanced && leftInfo.isBalanced;
        }
        if(rightInfo != null){
            isBalanced = isBalanced && rightInfo.isBalanced;
        }

        return new Info(height,isBalanced);
    }

    private class Info{
        private int height;
        private boolean isBalanced;

        public Info(int h,boolean is){
            height = h;
            isBalanced = is;
        }
    }
}