LeetCode – All Nodes Distance K in Binary Tree

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.

Note:

The given tree is non-empty.
Each node in the tree has unique values 0 <= node.val <= 500.
The target node is a node in the tree.
0 <= K <= 1000.

题意是让我们在一颗二叉树中，给定节点 Target, 寻找和target节点距离 为 K的所有节点，我们可以把这颗树看成一个无向图， 没有顺序就意味着，二叉树的旁边的指节也是可以算距离的。 
我们可以先把二叉树转化为无向图，再在图中进行 BFS 搜索应该就可以得到答案。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> distanceK(TreeNode root, TreeNode target, int K) {

        List<Integer> resList = new ArrayList<Integer> ();
        resList.add(target.val);

        Set<Integer> visited = new HashSet<>();
        visited.add(target.val);

        Map<Integer, List<Integer>> graph = new HashMap<>();
        treeToGraph(graph, null, root);

        if(graph.isEmpty() && target == root && K == 0){
            return resList;
        }
        else if(graph.isEmpty()){
            return new ArrayList<Integer> ();
        }

        for(int i=0; i<K; i++){
            List<Integer> temp = new ArrayList<>();
            for(Integer v : resList){
                for(Integer e : graph.get(v)){
                    if(!visited.contains(e)){
                        visited.add(e);
                        temp.add(e);
                    }
                }
            }
            resList = temp;
        }
        return resList;

    }

    public void treeToGraph(Map<Integer, List<Integer>> map, TreeNode parent, TreeNode child){
        if(parent != null && child != null){
            if(map.containsKey(parent.val)){
                map.get(parent.val).add(child.val);
            }
            else{
                List<Integer> list = new ArrayList<>();
                list.add(child.val);
                map.put(parent.val, list);
            }
            if(map.containsKey(child.val)){
                map.get(child.val).add(parent.val);
            }
            else{
                List<Integer> list = new ArrayList<>();
                list.add(parent.val);
                map.put(child.val, list);
            }
        }
        if(child.left != null){
            treeToGraph(map, child, child.left);
        }
        if(child.right != null){
            treeToGraph(map, child, child.right);
        }

    }
}