Binary Tree Nodes Around Radius - Amazon Top Interview Questions
Problem Statement :
You are given a binary tree root containing unique integers and integers target and radius. Return a sorted list of values of all nodes that are distance radius away from the node with value target. Constraints 1 ≤ n ≤ 100,000 where n is number of nodes in root 0 ≤ distance ≤ 100,000 Example 1 Input root = [3, [5, null, null], [2, [1, [6, null, null], [9, null, null]], [4, null, null]]] target = 4 radius = 2 Output [1, 3] Example 2 Input root = [0, null, null] target = 0 radius = 0 Output [0]
Solution :
Solution in C++ :
void fill(Tree* root, unordered_map<int, vector<int>>& graph) {
if (!root) return;
if (root->left) {
graph[root->val].push_back(root->left->val);
graph[root->left->val].push_back(root->val);
fill(root->left, graph);
}
if (root->right) {
graph[root->val].push_back(root->right->val);
graph[root->right->val].push_back(root->val);
fill(root->right, graph);
}
}
vector<int> solve(Tree* root, int target, int radius) {
unordered_map<int, vector<int>> graph;
fill(root, graph);
vector<int> q{target}, tmp;
if (!radius) return q;
unordered_set<int> seen{target};
while (!q.empty()) {
for (auto p : q) {
for (int c : graph[p]) {
if (seen.count(c)) continue;
seen.insert(c);
tmp.push_back(c);
}
}
q.clear(), q.swap(tmp);
radius--;
if (!radius) {
sort(q.begin(), q.end());
return q;
}
}
return q;
}
Solution in Python :
class Solution:
def solve(self, root, target, radius):
parent = defaultdict(lambda: None)
self.start = None
def postorder(root):
if not root:
return None
if root.val == target:
self.start = root
l, r = postorder(root.left), postorder(root.right)
parent[l] = parent[r] = root
return root
postorder(root)
# level order traversal radius levels
vis = set([None])
q = deque()
if self.start:
q.append(self.start)
cur_level = []
while q and radius >= 0:
cur_level_size = len(q)
cur_level = []
for _ in range(cur_level_size):
cur = q.popleft()
vis.add(cur)
cur_level.append(cur.val)
if parent[cur] not in vis:
q.append(parent[cur])
if cur.left not in vis:
q.append(cur.left)
if cur.right not in vis:
q.append(cur.right)
radius -= 1
return sorted(cur_level) if radius < 0 else []
View More Similar Problems
Left Rotation
A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d
View Solution →Sparse Arrays
There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun
View Solution →Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →