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[LeetCode][572] Subtree of Another Tree题解
题意：给一个树t和s，判断在t中是否有一个子树s。
思路：
dfs遍历一个树把t和s的val值一样的节点加入treenode*型的vector中，再使用搜索判断这个节点x开始，x和s是否一模一样，这里使用dfs来写isSame
代码
/*
* [572] Subtree of Another Tree
*
* https://leetcode.com/problems/subtree-of-another-tree/description/
*
* algorithms
* Easy (40.42%)
* Total Accepted: 70.4K
* Total Submissions: 174.1K
* Testcase Example: '[3,4,5,1,2]\n[4,1,2]'
*
*
* 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.
*
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
private:
vector<TreeNode *> res;
public:
bool isSubtree(TreeNode *s, TreeNode *t)
{
dfs(s, t);
for (int i = 0; i < res.size(); i++)
{
if (isSame(t, res[i]))
{
return true;
}
}
return false;
}
bool isSame(TreeNode *a, TreeNode *b)
{
if (a == nullptr && b == nullptr)
{
return true;
}
else if ((a == nullptr && b) || (a && b == nullptr))
{
return false;
}
else if (a->val == b->val)
{
return isSame(a->left, b->left) && isSame(a->right, b->right);
}
return false;
}
void dfs(TreeNode *s, TreeNode *t)
{
if (s)
{
if (s->val == t->val)
{
res.push_back(s);
}
if (s->left)
{
dfs(s->left, t);
}
if (s->right)
{
dfs(s->right, t);
}
}
}
};
提交结果：
√ Accepted
√ 176/176 cases passed (16 ms)
√ Your runtime beats 56.2 % of cpp submissions