给定一棵二叉树的头节点，判断是不是平衡二叉树

/**
 * 给定一棵二叉树的头节点，判断是不是平衡二叉树
 */
public class IsBalancedBT {

    public static boolean isBalancedBT(Node root) {
        if (root == null) {
            return true;
        }
        return process(root).isBalanced;
    }

    private static ResultInfo process(Node node) {
        if (node == null) {
            return new ResultInfo(0, true);
        }
        ResultInfo leftResult = process(node.left);
        ResultInfo rightResult = process(node.right);
        int height = Math.max(leftResult.height, rightResult.height) + 1;
        boolean isBalanced = true;
        if (!leftResult.isBalanced || !rightResult.isBalanced || Math.abs(leftResult.height - rightResult.height) > 1) {
            isBalanced = false;
        }
        return new ResultInfo(height, isBalanced);
    }

    /**
     * 向左右子树索要信息
     */
    public static class ResultInfo {

        // 当前高度
        public int height;

        // 是否平衡
        public boolean isBalanced;

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

    }

    /**
     * 二叉树结构
     */
    public static class Node {

        public Node left;

        public Node right;

    }

}

/* 如有意见或建议，欢迎评论区留言；如发现代码有误，欢迎批评指正 */