863. All Nodes Distance K in Binary Tree

https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/discuss/143752/JAVA-Graph-+-BFS

https://www.youtube.com/watch?v=o1siL8eKCos





class Solution {
    Map<TreeNode, List<TreeNode>> map = new HashMap();
//here can also use Map<TreeNode, TreeNode> to only store the child - parent mapping, since parent-child mapping is inherent in the tree structure
    
    public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {
         List<Integer> res = new ArrayList<Integer> ();
        if (root == null || K < 0) return res;
        buildMap(root, null); 
        if (!map.containsKey(target)) return res;
        Set<TreeNode> visited = new HashSet<TreeNode>();
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(target);
        visited.add(target);
        while (!q.isEmpty()) {
            int size = q.size();
            if (K == 0) {
                for (int i = 0; i < size ; i++) res.add(q.poll().val);
                return res;
            }
            for (int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                for (TreeNode next : map.get(node)) {
                    if(visited.add(next)){
                        q.offer(next);
                    }
                }
            }
            K--;
        }
        return res;
    }
    
    private void buildMap(TreeNode node, TreeNode parent) {
        if (node == null) return;
        if (!map.containsKey(node)) {
            map.put(node, new ArrayList<TreeNode>());
            if (parent != null)  { map.get(node).add(parent); map.get(parent).add(node) ; }
            buildMap(node.left, node);
            buildMap(node.right, node);
        }
    }    
}