Posted

Reading Time: 1 minute

We are given a binary tree (with root node root), a target node, and an integer value K.

Return a list of the values of all nodes that have a distance K from the target node.  The answer can be returned in any order.

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2

Output: [7,4,1]

Explanation: 
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.

Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Solution


/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
Map<Integer, List<Integer>> graph = new HashMap<>();


public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
buildGraph(null, root);

List<Integer> rst = new LinkedList<>();
Queue<Integer> queue = new LinkedList<>();
Set<Integer> visited = new HashSet<>();

int distance = 0;
queue.offer(target.val);
visited.add(target.val);

while(!queue.isEmpty()) {
int size = queue.size();
for(int i = 0; i < size; i++) {
int curr = queue.poll();
if(distance == K) {
rst.add(curr);
}

if(graph.containsKey(curr)) {
for(int neigh : graph.get(curr)) {
if(visited.contains(neigh)) {
continue;
}
queue.offer(neigh);
visited.add(neigh);
}
}
}
distance++;
}

return rst;
}

private void buildGraph(TreeNode parent, TreeNode child) {
if(parent != null) {
graph.computeIfAbsent(parent.val, key -> new LinkedList<>()).add(child.val);
graph.computeIfAbsent(child.val, key -> new LinkedList<>()).add(parent.val);
}

if(child.left != null) {
buildGraph(child, child.left);
}

if(child.right != null) {
buildGraph(child, child.right);
}
}
}