572. Subtree of Another Tree

最新推荐文章于 2024-09-25 18:14:06 发布

原创最新推荐文章于 2024-09-25 18:14:06 发布 · 214 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 同时被 2 个专栏收录

108 篇文章

订阅专栏

Leetcode_tree

15 篇文章

订阅专栏

本文探讨了如何确定一棵二叉树是否为另一棵二叉树的子结构。通过递归方法实现，首先检查主树是否存在与子树根节点相同的节点，然后验证该节点的左右子树是否分别与子树的左右子树匹配。

题目描述【Leetcode】

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:
…….3
….. / \
…. 4 5
…./ \
…1 2
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:
….3
…/ \
..4 5
./ \
1 2
.. /
..0
Given tree t:
…. 4
…./ \
…1 2
Return false.

判断某个树是否为另一个的子树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

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

    }
};