Print Nodes at distance K from a given node
You are given an arbitrary binary tree, a node of the tree, and an integer 'K'. You need to find all such nodes which have a distance K from the given node and return the list of these nodes.

Distance between two nodes in a binary tree is defined as the number of connections/edges in the path between the two nodes.

Note:

1. A binary tree is a tree in which each node has at most two children. 2. The given tree will be non-empty. 3. The given tree can have multiple nodes with the same value. 4. If there are no nodes in the tree which are at distance = K from the given node, return an empty list. 5. You can return the list of values of valid nodes in any order. For example if the valid nodes have values 1,2,3, then you can return {1,2,3} or {3,1,2} etc.

Input Format:

The first line of the input contains an integer T, denoting the number of test cases to run. The first line of each test case will contain the values of the nodes of the tree in the level order form ( -1 for NULL node). Refer to the example below for further explanation. The second line of each test case contains the value of the target node. The third and the last line of each test case contains the integer K denoting the distance at which nodes are to be found. Example: Consider the binary tree:

The input for the tree depicted in the above image would be : 3 5 1 6 2 0 8 -1 -1 7 4 -1 -1 -1 -1 -1 -1 -1 -1 Explanation : Level 1 : The root node of the tree is 3 Level 2 : Left child of 3 = 5 Right child of 3 = 1 Level 3 : Left child of 5 = 6 Right child of 5 = 2 Left child of 1 = 0 Right child of 1 = 8 Level 4 : Left child of 6 = null (-1) Right child of 6 = null(-1) Left child of 2 = 7 Right child of 2 = 4 Left child of 0 = null (-1) Right child of 0 = null (-1) Left child of 8 = null (-1) Right child of 8 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) Left child of 4 = null (-1) Right child of 4 = null (-1) The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null (-1).

Output Format :

For each test case, print the values of all nodes at distance = K, from the given target node.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Print Nodes at distance K from a given node

You are given an arbitrary binary tree, a node of the tree, and an integer 'K'. You need to find all such nodes which have a distance K from the given node and return the list of these nodes.

Distance between two nodes in a binary tree is defined as the number of connections/edges in the path between the two nodes.

Note:

1. A binary tree is a tree in which each node has at most two children. 
2. The given tree will be non-empty.
3. The given tree can have multiple nodes with the same value.
4. If there are no nodes in the tree which are at distance = K from the given node, return an empty list.
5. You can return the list of values of valid nodes in any order. For example if the valid nodes have values 1,2,3, then you can return {1,2,3} or {3,1,2} etc.

Input Format:

The first line of the input contains an integer T, denoting the number of test cases to run. 

The first line of each test case will contain the values of the nodes of the tree in the level order form ( -1 for NULL node). Refer to the example below for further explanation.

The second line of each test case contains the value of the target node.

The third and the last line of each test case contains the integer K denoting the distance at which nodes are to be found. 

Example:

Consider the binary tree:

Input Format Fig.

The input for the tree depicted in the above image would be :

3
5 1
6 2 0 8
-1 -1 7 4 -1 -1 -1 -1
-1 -1 -1 -1

Explanation :
Level 1 :
The root node of the tree is 3

Level 2 :
Left child of 3 = 5
Right child of 3 = 1

Level 3 :
Left child of 5 = 6
Right child of 5 = 2
Left child of 1 = 0
Right child of 1 = 8

Level 4 :
Left child of 6 = null (-1)
Right child of 6 = null(-1)
Left child of 2 = 7
Right child of 2 = 4
Left child of 0 = null (-1)
Right child of 0 = null (-1)
Left child of 8 = null (-1)
Right child of 8 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
Left child of 4 = null (-1)
Right child of 4 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).

Output Format :

For each test case, print the values of all nodes at distance = K, from the given target node.

Note:

You do not need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 100
1 <= N <= 3000
0 <= K <= 3000
0 <= nodeValue <= 3000

Where nodeValue donates the value of the node.

Time Limit: 1 sec

Accepted Answer

Tip 1:We can solve this problem using breadth first search. Main thing to observe in this problem is that if we find two marked nodes which are at largest distance from each other considering all pairs of marked nodes then if a node is at a distance less than K from both of these two nodes then it will be at a distance less than K from all the marked nodes because these two nodes represents the extreme limit of all marked nodes, if a node lies in this limit then it will be at a distance less than K from all marked nodes otherwise not.
As in above example, node-1 and node-4 are most distant marked node so nodes which are at distance less than 3 from these two nodes will also be at distance less than 3 from node 2 also. Now first distant marked node we can get by doing a bfs from any random node, second distant marked node we can get by doing another bfs from marked node we just found from the first bfs and in this bfs we can also found distance of all nodes from first distant marked node and to find distance of all nodes from second distant marked node we will do one more bfs, so after doing these three bfs we can get distance of all nodes from two extreme marked nodes which can be compared with K to know which nodes fall in K-distance range from all marked nodes.

Accepted Answer

DFS

Create a map to store the parent of each node in the tree, Traverse the tree recursively (via depth-first search), at each step if the current node is not NULL. Store its parent in the map, then traverse the left and right subtree.
Now assume that the given node is the root of the tree. In such a case, we can simply run a breadth-first search from the root, and track the current level of the tree. When the level = ‘K’, then the current nodes in the queue will be our answer.
But the problem is if the target node is not the root, then we can’t travel to every node with only left and right pointers. So here the stored parents will help us.
Observe that in a binary tree a node can be connected to maximum 3 other nodes ie. left child, right child, and the parent. We have left and right pointers, and we have also stored the parent node.
Now simply start BFS from the target node, and for each node which is at front of the queue, push its left child, right child and parent node(provided they are not null).
When the level of BFS reaches ‘K’, break the BFS and store all the values of nodes in the queue to an array and return it.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given tree.

As we use additional space in BFS for the queue, and in the recursion call stack for DFS. Hence the overall space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given tree.

As we are checking for every node once, hence the time complexity will be O(N).

Accepted Answer

Insertion with Depth Calculation.

A general idea for this approach is that if from the root, the target lies in its left branch at distance ‘X’, then all nodes in its right branch at distance ‘K’ - ‘X’, should be included in the answer.
Traverse every node via DFS, DFS(node) will return us the distance between node and target.
If ‘NODE’ == ‘TARGET’, then we should add all such nodes to our answer that are at distance ‘K’ from the target in the subtree of the target.
If the target is in the left branch of the node, assume at distance ‘X’, then add all such nodes at distance ‘K’ - ‘X’ in the right branch. If ‘K’ - ‘X’ depth doesn’t exist in the right branch, return.
And similarly, do if the target lies in the left branch.
If the target doesn’t lie in both left and right branches, then stop.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given tree.

We are using a stack of size ‘N’ for recursion. Hence the overall space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given tree.

As we are checking for every node once, hence the time complexity will be O(N).

You are given an arbitrary binary tree, a node of the tree, and an integer 'K'. You need to find all such nodes which have a distance K from the given node and return the list of these nodes.

Distance between two nodes in a binary tree is defined as the number of connections/edges in the path between the two nodes.

Note:

Input Format:

Output Format :

Note:

Constraints:

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