[Leetcode] 572. Subtree of Another Tree 解题报告

题目：

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.

思路：

练手题目，哈哈。

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) {        vector<TreeNode*> nodes;        preorder(s, nodes);        for (auto n : nodes) {            if (sameTree(t, n)) {                return true;            }        }        return false;    }private:    void preorder(TreeNode *s, vector<TreeNode*> &nodes) {        if (s == NULL) {            return;        }        nodes.push_back(s);        preorder(s->left, nodes);        preorder(s->right, nodes);    }    bool sameTree(TreeNode *t1, TreeNode *t2) {        if (t1 == NULL && t2 == NULL) {            return true;        }        else if (t1 == NULL || t2 == NULL) {            return false;        }        else {            if (t1->val != t2->val) {                return false;            }            else {                return sameTree(t1->left, t2->left) && sameTree(t1->right, t2->right);            }        }    }};