LeetCode 572. Subtree of Another Tree

最新推荐文章于 2021-05-02 17:19:34 发布

WX_ming

最新推荐文章于 2021-05-02 17:19:34 发布

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本文链接：https://blog.youkuaiyun.com/WX_ming/article/details/81320905

本文探讨了如何判断一棵树是否为另一棵树的子树。通过详细的代码解释，展示了如何使用递归方法比较两棵树的结构和节点值，以确定子树的存在。提供了两个示例，分别说明了当子树存在和不存在时的判断过程。

572. Subtree of Another Tree

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

Given tree t:

     4 
   /   \
  1     2

Return true, because t has the same structure and node values with a subtree of s.
Example 2:
Given tree s:

Given tree t:

  4
 / \
1   2

Return false.

Approach

题目大意：给你树t和树s，问你是否s是t的子树
思路方法：枚举每个结点的结构是否与s相同

Code

class Solution {
public:
    bool isSubtree(TreeNode* s, TreeNode* t) {
        return traval(s, t);
    }
    bool equal(TreeNode* s,TreeNode* t){
        if (s == nullptr&&t == nullptr)return true;
        if (s == nullptr&&t != nullptr)return false;
        if (s != nullptr&&t == nullptr)return false;
        return (s->val == t->val) && equal(s->left, t->left) && equal(s->right, t->right);
    }
    bool traval(TreeNode* s, TreeNode* t) {
        return s != nullptr && (equal(s, t) || traval(s->left,t)||traval(s->right,t));
    }