剑指offer系列43---判断平衡二叉树

Posted 2020-07-28 noaman_wgs

【题目】判断一颗二叉树是不是平衡二叉树。
* 平衡二叉树定义：任意子节点深度相差不超过1.
【思路】由上题，利用递归得到二叉树每个结点的深度同时比较。

 1 package com.exe9.offer;
 2 
 3 import com.exe9.offer.BTreeDepth.TreeNode;
 4 
 5 /**
 6  * 【题目】判断一颗二叉树是不是平衡二叉树。
 7  *           平衡二叉树定义：任意子节点深度相差不超过1.
 8  * 【思路】由上题，利用递归得到二叉树每个结点的深度同时比较。
 9  * @author WGS
10  *
11  */
12 public class IsAVL {
13 
14     static class TreeNode{
15         int val;
16         TreeNode left=null;
17         TreeNode right=null;
18         public TreeNode(int val){
19             this.val=val;
20         }
21     }
22     //得到树的深度
23     public int getBTreeDepth(TreeNode pRoot){
24         if(pRoot==null) return 0;
25         //每当proot==null时，就比较left与right大小，加1
26         int left=getBTreeDepth(pRoot.left);
27         int right=getBTreeDepth(pRoot.right);
28         //System.out.println(left+"===="+right);
29         return (left>right)?left+1:right+1;
30         
31     }
32     public boolean isAVL(TreeNode pRoot){
33         if(pRoot==null) return true;
34         int left=getBTreeDepth(pRoot.left);
35         int right=getBTreeDepth(pRoot.right);
36         int val=left-right;
37         if(val>1 || val<-1)
38             return false;
39         
40         return isAVL(pRoot.left) && isAVL(pRoot.right);
41     }
42     public static void main(String[] args) {
43         TreeNode root = new TreeNode(1);
44         TreeNode node1 = new TreeNode(2);
45         TreeNode node2 = new TreeNode(3);
46         TreeNode node3 = new TreeNode(4);
47         TreeNode node4 = new TreeNode(5);
48         TreeNode node5 = new TreeNode(6);
49         TreeNode node6 = new TreeNode(7);
50         
51         root.left = node1;
52         root.right = node2;
53         node1.left = node3;
54         node1.right = node4;
55         node2.right = node5;
56         node4.left = node6;
57         
58         boolean b=new IsAVL().isAVL(root);
59         System.out.println(b);
60 
61     }
62 
63 }