Leetcode算法学习日志-572 Subtree of Another Tree

Leetcode 572 Subtree of Another Tree

题目原文

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.

题意分析

判断树t是不是树s的子树。

解法分析

该题采用递归的思想解决，比较树s和树t，如果s=t，则t一定是s的子树，不然递归s的两个子树，判断t是不是这两个子树的子树，是其中一个的子树则函数返回1.C++代码如下
/**
* 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 theSame(TreeNode* s, TreeNode*t){
if(t==NULL||s==NULL){//This is important
if(t==s)
return true;
else
return false;
}
if((s->val)!=(t->val))
return false;
return theSame(s->left,t->left)&&theSame(s->right,t->right);
}
bool isSubtree(TreeNode* s, TreeNode* t) {
if(theSame(s,t))
return true;
if(s==NULL)//This is important
return false;
return isSubtree(s->left,t)||isSubtree(s->right,t);
}
};上述代码需要注意对空指针的处理，不然会造成runtime error。