# Subtree - Amazon Top Interview Questions

### Problem Statement :

```You are given two binary trees root, and target. Return whether target is a subtree of root — that is, whether there's a node in root that is identically same in values and structure as root including all of its descendants.

Example 1

Input

root = [1, [2, null, null], [3, [4, [6, null, null], null], [5, null, [7, null, null]]]]
target = [3, [4, [6, null, null], null], [5, null, [7, null, null]]]

Output

True

Example 2

Input

root = [1, [2, null, null], [3, [4, [6, null, null], null], [5, null, [7, null, null]]]]
target = [3, null, [5, null, null]]

Output

False

Example 3

Input

root = [0, null, [5, [1, null, null], null]]
target = [0, null, [5, [1, null, null], null]]

Output

True```

### Solution :

```                        ```Solution in C++ :

/**
* class Tree {
*     public:
*         int val;
*         Tree *left;
*         Tree *right;
* };
*/
bool flag;
bool identical(Tree* root, Tree* target) {
if (!root && !target) return true;
if (!root || !target) return false;
if (root->val != target->val) return false;
return identical(root->left, target->left) && identical(root->right, target->right);
}
void traverse(Tree* root, Tree* target) {
if (!root) return;
if (root->val == target->val) {
flag = flag | identical(root, target);
}
traverse(root->left, target);
traverse(root->right, target);
}
bool solve(Tree* root, Tree* target) {
flag = false;
if (!target) return true;
traverse(root, target);
return flag;
}```
```

```                        ```Solution in Python :

# class Tree:
#     def __init__(self, val, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def merkleHash(self, root):
if not root:
return "-"

leftHash = self.merkleHash(root.left)
rightHash = self.merkleHash(root.right)

root.merkle = leftHash + ":{}:".format(root.val) + rightHash
return root.merkle

def solve(self, root, target):
self.merkleHash(root)
self.merkleHash(target)

def dfs(cur):
if not cur:
return False

return cur.merkle == target.merkle or dfs(cur.left) or dfs(cur.right)

return dfs(root)```
```

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