判断一棵二叉树是否是平衡二叉树

思路

递归判断每个节点及其子节点组成的树是否是平衡树

代码

package BinaryTree;

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

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

    public static class ReturnData{
        public boolean isB;
        public int level;

        public ReturnData(boolean isB,int data){
            this.isB=isB;
            this.level=data;
        }
    }

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

    public static ReturnData process(Node head){
        if (head==null){
            return new ReturnData(true,0);
        }
        ReturnData LeftReturnData=process(head.left);
        if (LeftReturnData.isB==false){
            return new ReturnData(false,0);
        }
        ReturnData RightReturnData=process(head.right);
        if(RightReturnData.isB==false){
            return new ReturnData(false,0);
        }
        if (Math.abs(LeftReturnData.level-RightReturnData.level)>1){
            return new ReturnData(false,0);
        }
        return new ReturnData(true,Math.max(LeftReturnData.level,RightReturnData.level)+1);
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        head.left = new Node(2);
        head.right = new Node(3);
        head.left.left = new Node(4);
        head.left.right = new Node(5);
        head.right.left = new Node(6);
        head.right.right = new Node(7);

        System.out.println(isBalance(head));

    }

}