树的子结构判断

 1 /**
 2 public class TreeNode {
 3     int val = 0;
 4     TreeNode left = null;
 5     TreeNode right = null;
 6 
 7     public TreeNode(int val) {
 8         this.val = val;
 9 
10     }
11 
12 }
13 */
14 public class Solution {
15     //判断root2是否是root1的子树
16     public boolean HasSubtree(TreeNode root1,TreeNode root2) {
17         if(root1 == null || root2 == null)
18             return false;
19         if(isEqualTree(root1,root2))
20             return true;
21         else
22             return HasSubtree(root1.left,root2) || HasSubtree(root1.right,root2);        
23     }
24     //判断B是否是A的子树，并且从跟节点开始
25     public boolean isEqualTree(TreeNode A,TreeNode B){
26         if(B == null) return true;
27         if(A == null) return false;
28         if(A.val != B.val)return false;
29         return isEqualTree(A.left,B.left) && isEqualTree(A.right,B.right);
30         
31     }
32 }