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/**

* 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.

*

* https://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/sketch0.png

*

* 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.

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class AllNodesDistanceKInBinaryTree863 {

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

List<Integer> res = new ArrayList<>();

if (root == null) return res;

if (K == 0) {

res.add(target.val);

return res;

}

distanceToRoot(root, target, K, res, 0);

return res;

}

private int distanceToRoot(TreeNode root, TreeNode target, int K, List<Integer> res, int level) {

if (root == null) return -1;

if (root.val == target.val) {

distanceKChildren(root, K, res);

return level;

}

int leftFlag = distanceToRoot(root.left, target, K, res, level + 1);

if (leftFlag != -1) {

if (leftFlag - level == K) {

res.add(root.val);

} else if (leftFlag - level < K) {

distanceKChildren(root.right, K - (leftFlag - level) - 1, res);

}

}

int rightFlag = distanceToRoot(root.right, target, K, res, level + 1);

if (rightFlag != -1) {

if (rightFlag - level == K) {

res.add(root.val);

} else if (rightFlag - level < K) {

distanceKChildren(root.left, K - (rightFlag - level) - 1, res);

}

}

return leftFlag != -1 ? leftFlag : rightFlag;

}

private void distanceKChildren(TreeNode root, int K, List<Integer> res) {

if (root == null) return;

Queue<TreeNode> q = new LinkedList<>();

q.add(root);

int i = 0;

while (!q.isEmpty() && i < K) {

int size = q.size();

for (int j=0; j<size; j++) {

TreeNode curr = q.poll();

if (curr.left != null) q.add(curr.left);

if (curr.right != null) q.add(curr.right);

}

i++;

}

while (!q.isEmpty()) {

res.add(q.poll().val);

}

}

/**

* https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/solution/

*/

Map<TreeNode, TreeNode> parent;

public List<Integer> distanceK2(TreeNode root, TreeNode target, int K) {

parent = new HashMap();

dfs(root, null);

Queue<TreeNode> queue = new LinkedList();

queue.add(null);

queue.add(target);

Set<TreeNode> seen = new HashSet();

seen.add(target);

seen.add(null);

int dist = 0;

while (!queue.isEmpty()) {

TreeNode node = queue.poll();

if (node == null) {

if (dist == K) {

List<Integer> ans = new ArrayList();

for (TreeNode n: queue)

ans.add(n.val);

return ans;

}

queue.offer(null);

dist++;

} else {

if (!seen.contains(node.left)) {

seen.add(node.left);

queue.offer(node.left);

}

if (!seen.contains(node.right)) {

seen.add(node.right);

queue.offer(node.right);

}

TreeNode par = parent.get(node);

if (!seen.contains(par)) {

seen.add(par);

queue.offer(par);

}

}

}

return new ArrayList<Integer>();

}

public void dfs(TreeNode node, TreeNode par) {

if (node != null) {

parent.put(node, par);

dfs(node.left, node);

dfs(node.right, node);

}

}

/**

* https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/solution/

*/

List<Integer> ans;

TreeNode target;

int K;

public List<Integer> distanceK3(TreeNode root, TreeNode target, int K) {

ans = new LinkedList();

this.target = target;

this.K = K;

dfs(root);

return ans;

}

// Return distance from node to target if exists, else -1

public int dfs(TreeNode node) {

if (node == null)

return -1;

else if (node == target) {

subtree_add(node, 0);

return 1;

} else {

int L = dfs(node.left), R = dfs(node.right);

if (L != -1) {

if (L == K) ans.add(node.val);

subtree_add(node.right, L + 1);

return L + 1;

} else if (R != -1) {

if (R == K) ans.add(node.val);

subtree_add(node.left, R + 1);

return R + 1;

} else {

return -1;

}

}

}

// Add all nodes 'K - dist' from the node to answer.

public void subtree_add(TreeNode node, int dist) {

if (node == null) return;

if (dist == K)

ans.add(node.val);

else {

subtree_add(node.left, dist + 1);

subtree_add(node.right, dist + 1);

}

}

Map<TreeNode, Integer> map = new HashMap<>();

public List<Integer> distanceK4(TreeNode root, TreeNode target, int K) {

List<Integer> res = new LinkedList<>();

find(root, target, K);

dfs(root, target, K, map.get(root), res);

return res;

}

// find target node first and store the distance in that path that we could use it later directly

private int find(TreeNode root, TreeNode target, int K) {

if (root == null) return -1;

if (root == target) {

map.put(root, 0);

return 0;

}

int left = find(root.left, target, K);

if (left >= 0) {

map.put(root, left + 1);

return left + 1;

}

int right = find(root.right, target, K);

if (right >= 0) {

map.put(root, right + 1);

return right + 1;

}

return -1;

}

private void dfs(TreeNode root, TreeNode target, int K, int length, List<Integer> res) {

if (root == null) return;

if (map.containsKey(root)) length = map.get(root);

if (length == K) res.add(root.val);

dfs(root.left, target, K, length + 1, res);

dfs(root.right, target, K, length + 1, res);

}

}