Subtree of Another Tree问题及解法

问题描述：

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.

示例：

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.

Given tree s:

3
/ \
4   5
/ \
1   2/
0

Given tree t:

4
/ \
1   2

Return false.
问题分析：

该问题与之前的两棵树是否相同的题目原理一样，我在此处采用了BFS，利用队列求解。

过程详见代码：

/**
* 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 {
public:
bool isSubtree(TreeNode* s, TreeNode* t) {
if(s == NULL && t != NULL) return false;
queue<TreeNode* > sq;
queue<TreeNode* > tq;
TreeNode* tt = t;
TreeNode* ss = s;
if(s->val == t->val)
{
if(t->left != NULL) tq.push(t->left);
if(t->right != NULL)tq.push(t->right);
if(s->left != NULL)sq.push(s->left);
if(s->right != NULL)sq.push(s->right);

while(!tq.empty() && !sq.empty())
{
t = tq.front();
s = sq.front();
tq.pop();
sq.pop();
if(s->val == t->val)
{
if(t->left != NULL) tq.push(t->left);
if(t->right != NULL)tq.push(t->right);
if(s->left != NULL)sq.push(s->left);
if(s->right != NULL)sq.push(s->right);
}
else return isSubtree(ss->left,tt) || isSubtree(ss->right,tt);
}
return tq.empty() && sq.empty() || isSubtree(ss->left,tt) || isSubtree(ss->right,tt);
}
return isSubtree(ss->left,tt) || isSubtree(ss->right,tt);
}
};

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