leetcode 572. Subtree of Another Tree 子树判断 + 深度优先遍历DFS

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.

本题题意很简单，直接递归遍历即可

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>

using namespace std;

/*
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution
{
public:
bool isSubtree(TreeNode* s, TreeNode* t)
{
if (s == NULL)
return false;
else if (isSame(s, t) == true)
return true;
else
return isSubtree(s->left, t) || isSubtree(s->right, t);
}

bool isSame(TreeNode* s, TreeNode* t)
{
if (s == NULL && t == NULL)
return true;
else if (s == NULL && t != NULL || s != NULL && t == NULL)
return false;
else
{
if (s->val != t->val)
return false;
else
return isSame(s->left,t->left) && isSame(s->right,t->right);
}
}
};