1 public class Solution {
 2     public static boolean HasSubtree(TreeNode root1, TreeNode root2) {
 3         boolean result = false;
 4         //当Tree1和Tree2都不为零的时候，才进行比较。否则直接返回false
 5         if (root2 != null && root1 != null) {
 6             //如果找到了对应Tree2的根节点的点
 7             if(root1.val == root2.val){
 8                 //以这个根节点为为起点判断是否包含Tree2
 9                 result = doesTree1HaveTree2(root1,root2);
10             }
11             //如果找不到，那么就再去root的左孩子当作起点，去判断时候包含Tree2
12             if (!result) {
13                 result = HasSubtree(root1.left,root2);
14             }
15              
16             //如果还找不到，那么就再去root的右孩子当作起点，去判断时候包含Tree2
17             if (!result) {
18                 result = HasSubtree(root1.right,root2);
19                }
20             }
21             //返回结果
22         return result;
23     }
24  
25     public static boolean doesTree1HaveTree2(TreeNode node1, TreeNode node2) {
26         //如果Tree2已经遍历完了都能对应的上，返回true
27         if (node2 == null) {
28             return true;
29         }
30         //如果Tree2还没有遍历完，Tree1却遍历完了。返回false
31         if (node1 == null) {
32             return false;
33         }
34         //如果其中有一个点没有对应上，返回false
35         if (node1.val != node2.val) {   
36                 return false;
37         }
38          
39         //如果根节点对应的上，那么就分别去子节点里面匹配
40         return doesTree1HaveTree2(node1.left,node2.left) && doesTree1HaveTree2(node1.right,node2.right);
41     }
42 }