**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 []
```

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