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.


1 ≤ n ≤ 100,000 where n is number of nodes in root
0 ≤ distance ≤ 100,000

Example 1


root = [3, [5, null, null], [2, [1, [6, null, null], [9, null, null]], [4, null, null]]]
target = 4
radius = 2


[1, 3]

Example 2


root = [0, null, null]
target = 0
radius = 0



Solution :


                        Solution in C++ :

void fill(Tree* root, unordered_map<int, vector<int>>& graph) {
    if (!root) return;
    if (root->left) {
        fill(root->left, graph);
    if (root->right) {
        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;
        q.clear(), q.swap(tmp);
        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

        # level order traversal radius levels
        vis = set([None])
        q = deque()
        if 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()

                if parent[cur] not in vis:
                if cur.left not in vis:
                if cur.right not in vis:

            radius -= 1

        return sorted(cur_level) if radius < 0 else []

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