什么是平衡二叉树（BBT）？

1. |当前节点的左子树高度-当前节点的右子树高度| <= 1
2. 左子树和右子树也是平衡二叉树

代码实现：

public class BalancedTree {
    public static class Node {
        public int value;
        public Node left;
        public Node right;

        public Node(int value) {
            this.value = value;
        }
    }

    // 基于上述要求，说明每个节点调用递归函数中需要返回两个信息：左右子树是否平衡以及高度大大小
    static class ReturnType {
        public boolean isBalanced;
        public int height;

        public ReturnType(boolean isBalanced, int height) {
            this.isBalanced = isBalanced;
            this.height = height;
        }
    }

    public static boolean isBalanced(Node head) {
        return process(head).isBalanced;
    }

    public static ReturnType process(Node x) {
        // base case
        if (x == null) {
            return new ReturnType(true, 0);
        }

        // 1. 分别对左树和右树进行处理，返回需要的两个信息(即ReturnType对象)
        ReturnType leftData = process(x.left);
        ReturnType rightData = process(x.right);

        // 2. 计算以当前节点为根节点的树的高度
        int height = Math.max(leftData.height, rightData.height) + 1;

        // 3. 判断以当前节点为根节点的树是否为平衡二叉数(即判断是否满足上述两个条件)
        boolean isBalanced = leftData.isBalanced &&
                             rightData.isBalanced &&
                             Math.abs(leftData.height - rightData.height) <= 1;

        // 4. 将得到的两个信息封装为对象返回
        return new ReturnType(isBalanced, height);
    }
}