2019-11-02 12:07:41 +08:00
|
|
|
# 26. 树的子结构
|
|
|
|
|
|
|
|
|
|
[NowCoder](https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
|
|
|
|
|
|
|
|
|
## 题目描述
|
|
|
|
|
|
2019-11-02 17:33:10 +08:00
|
|
|
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/84a5b15a-86c5-4d8e-9439-d9fd5a4699a1.jpg" width="450"/> </div><br>
|
2019-11-02 12:07:41 +08:00
|
|
|
|
|
|
|
|
## 解题思路
|
|
|
|
|
|
|
|
|
|
```java
|
|
|
|
|
public boolean HasSubtree(TreeNode root1, TreeNode root2) {
|
|
|
|
|
if (root1 == null || root2 == null)
|
|
|
|
|
return false;
|
|
|
|
|
return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
|
|
|
|
|
if (root2 == null)
|
|
|
|
|
return true;
|
|
|
|
|
if (root1 == null)
|
|
|
|
|
return false;
|
|
|
|
|
if (root1.val != root2.val)
|
|
|
|
|
return false;
|
|
|
|
|
return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
|
|
|
|
|
}
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2019-11-02 17:33:10 +08:00
|
|
|
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
