剑指Offer 26 树的子结构

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

 1 # -*- coding:utf-8 -*-
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 class Solution:
 8     def isSubWithRoot(self,pRoot1,pRoot2):
 9         if pRoot2 == None:
10             return True
11         if pRoot1 == None:
12             return False
13         if pRoot1.val != pRoot2.val:
14             return False
15         return self.isSubWithRoot(pRoot1.left,pRoot2.left) and self.isSubWithRoot(pRoot1.right,pRoot2.right)
16     
17     def HasSubtree(self, pRoot1, pRoot2):
18         if pRoot1 == None or pRoot2 == None:
19             return False
20         return self.isSubWithRoot(pRoot1,pRoot2) or self.HasSubtree(pRoot1.left,pRoot2) or self.HasSubtree(pRoot1.right,pRoot2)
21         # write code here

Java版本，leetcode地址：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public boolean isSubStructure(TreeNode A, TreeNode B) {
12         if(A == null || B == null){
13             return false;
14         }
15         return isSubTree(A,B) || isSubStructure(A.left,B) || isSubStructure(A.right,B);
16     }
17 
18     public boolean isSubTree(TreeNode A,TreeNode B){
19         if(B == null){
20             return true;
21         }
22         if(A == null){
23             return false;
24         }
25         if(A.val != B.val){
26             return false;
27         }
28         return isSubTree(A.left,B.left) && isSubTree(A.right,B.right);
29     }
30 }

posted on 2019-06-12 22:15 Sempron2800+ 阅读(92) 评论(0) 编辑收藏举报