# Leetcode [No. 572] Subtree of Another Tree (EASY)

## 題目

https://leetcode.com/problems/subtree-of-another-tree/description/

## 思路
+ 這個解法主要是把問題拆為兩個步驟：
+ 第一個步驟是先去traverse這顆tree，直到他們的root是相同的
+ 第二個步驟是當他們的頭相同後，我們再去比較他們的子樹是否相同
+ 如果不相同的話，我們就去判斷root的子樹有沒有可能會成功這樣～

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* root, TreeNode* subRoot) 
    {
        // if subRoot is null, it must be the subtree of the root.
        if(subRoot == nullptr) 
        {
            return true;
        }
        // but if root is null, it would not have any subtree.
        if(root == nullptr)
        {
            return false;
        }
        if(isSameTree(root, subRoot))
        {
            return true;
        }
        // else traverse to the left node and right node
        return isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
        
    }
    bool isSameTree(TreeNode* root, TreeNode* subRoot)
    {
        if(root == subRoot) // both of them are nullptr
        {
            return true;
        }
        else if(!root && subRoot || root && !subRoot)
        {
            return false;
        }
        return root->val == subRoot->val && isSameTree(root->left, subRoot->left) && isSameTree(root->right, subRoot->right);
    }
};

### 解法分析
+ time complexity: O(n*m)

### 執行結果

---

## 改良

+ DFS裡面的function很重要，因為只要有一個node是true就是true，所以要用OR來return.

class Solution {
public:
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        return dfs(root, subRoot);
    }

    bool dfs(TreeNode* root, TreeNode* subRoot) {
        if (!root) return false;
        if (root->val == subRoot->val && isSameTree(root, subRoot)) {
            return true;
        }
        return dfs(root->left, subRoot) || dfs(root->right, subRoot);
    }

    // copy from q100.
    bool isSameTree(TreeNode* p, TreeNode* q) {
        if (p == q) return true; // both null
        if (p && q) {
            return p->val == q->val && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
        } else {
            // one is null and other not
            return false;
        }
    }
};