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.
分析：先把tree转成graph，然后bfs。
1 class Solution { 2 Map<TreeNode, List<TreeNode>> map = new HashMap(); 3 4 public List<Integer> distanceK(TreeNode root, TreeNode target, int K) { 5 List<Integer> res = new ArrayList<> (); 6 if (root == null || K < 0) return res; 7 8 buildGraphFromTree(root, null); 9 if (!map.containsKey(target)) return res; 10 11 Set<TreeNode> visited = new HashSet<>(); 12 Queue<TreeNode> q = new LinkedList<>(); 13 q.add(target); 14 visited.add(target); 15 while (!q.isEmpty()) { 16 int size = q.size(); 17 if (K == 0) { 18 while (!q.isEmpty()) { 19 res.add(q.poll().val); 20 } 21 return res; 22 } 23 for (int i = 0; i < size; i++) { 24 TreeNode node = q.poll(); 25 for (TreeNode neighbor : map.get(node)) { 26 if (!visited.contains(neighbor)) { 27 visited.add(neighbor); 28 q.add(neighbor); 29 } 30 } 31 } 32 K--; 33 } 34 return res; 35 } 36 37 private void buildGraphFromTree(TreeNode node, TreeNode parent) { 38 if (node != null) { 39 map.put(node, new ArrayList<TreeNode>()); 40 if (parent != null) { 41 map.get(node).add(parent); 42 map.get(parent).add(node); 43 } 44 buildGraphFromTree(node.left, node); 45 buildGraphFromTree(node.right, node); 46 } 47 } 48 }
All Nodes Distance K in Binary Tree
原文：https://www.cnblogs.com/beiyeqingteng/p/12254338.html