[LeetCode]572. Subtree of Another Tree

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 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.

---

/**
 * 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) {
        bool result = false;
        // 保证不为空的情况下才传给子函数处理，t题目其实已经说了不为空了
        if(s != NULL && t != NULL){
            // 如果值相同，则调用is函数判断是不是相同的子树
            if(s->val == t->val)
                result = is(s, t);
            // 如果根结点不是，则分别递归判断左右子结点
            if(result == false)
                result = isSubtree(s->left, t);
            if(result == false)
                result = isSubtree(s->right, t);
        }
        return result;
    }

    // 此函数判断root1和root2 是不是相同的子树
    bool is(TreeNode* root1, TreeNode* root2){
        // 题目强调的是子树，不是子结构，所以只有遍历到都为空时才返回true
        if(root1 == NULL && root2 == NULL)
            return true;
        if(root1 == NULL || root2 == NULL)
            return false;
        if(root1->val != root2->val)
            return false;
        return is(root1->left, root2->left) && is(root1->right, root2->right);
    }
};