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


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]]]



Example 2


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



Example 3


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



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):

        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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