LeetCode

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.

居然卡题了，挺蠢的。。。本来想的是先找到值相等的地方，再开始进行循环判断，结果发现多个值跟t->val相等的时候就会判断不出来，就，很难受。。。

然后干脆直接写循环吧。。。行吧。。。

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

赋上WA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 (t == NULL) return true;        if (s == NULL) return false;        TreeNode* s1 = s;        stack<TreeNode*> sta;        sta.push(s);        while (true && !sta.empty()) {            TreeNode* cur = sta.top();            sta.pop();            if (cur->val == t->val) {                s = cur;                break;            }            if (cur->left) sta.push(cur->left);            if (cur->right) sta.push(cur->right);        }        bool ans = solve(s, t);        return ans;    }    bool solve(TreeNode* s, TreeNode* t) {        if (s == NULL && t == NULL) return true;        if (s == NULL || t == NULL) return false;        if (s->val != t->val) return false;        return solve(s->left, t->left) && solve(s->right, t->right);    }};