【LeetCode & 剑指offer刷题】树题8：26 树的子结构（572. Subtree of Another Tree）
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
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 scould 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) {}
* };
*/
/*
问题：判断某子树是否是主树的子结构
方法：前序遍历的递归方法
主树s， 子树t
*/
class Solution
{
public:
bool isSubtree(TreeNode* s, TreeNode* t) //遍历主树s（前序遍历递归法）
{
if(s == nullptr) return false; //前序遍历递归的出口，注意一定要加判断空指针的语句，后面有s->left和s->right
//前序遍历（递归法）主树t各结点,从根结点到左子树再到右子树（s一直在延伸，展开分支）
if(isSame(s,t)) //判断当前结点下的子树是否一样
return true;
else //判断当前结点的左子树是否为子树t,再判断右子树是否为子树t
return isSubtree(s->left, t) || isSubtree(s->right, t);
}
private:
bool isSame(TreeNode* s, TreeNode* t) //确定s父结点后，开始同时扫描（递归法）s和t，看是各个结点是否相等
{
if(s == nullptr && t == nullptr) return true; //如果最后均遍历到空结点，返回true
else if(s == nullptr || t == nullptr) return false; //如果一个遍历到空，一个没有，说明不同，返回false
if(s->val == t->val)
{
return isSame(s->left, t->left) && isSame(s->right, t->right);
}
else
return false; //不相同，返回假
}
};
