**Problem**

You are given a function printKDistanceNodes which takes in a root
node of a binary tree, a start node and an integer K. Complete the
function to print the value of all the nodes (one-per-line) which are a K
distance from the given start node in sorted order. Distance can be
upwards or downwards.

start node = 18, k = 2 , then output = 2, 19, 25

**Example**start node = 18, k = 2 , then output = 2, 19, 25

start node = 18, k = 3, then output = -4, 3

### Solution

We have already seen a similar problem, where we have to find k distance from the root and k distance from the leaf. Find the distance from root is easy. In the second case of printing from bottom to top (k distance from leaves), we know the direction, i.e. we have to go up. But here we have to find the k elements even going upwards.Note :- Parent pointer is not given.

**Method 1 - Using the recursion**

(Printing nodes at a disance of K downwards is easy). Its a simple recursive function.So moving to nodes which are in upwards direction.

There are two types of nodes to be considered.

**1)**Nodes in the subtree rooted with target node. For example if the target node is 18 and k is 2, then such nodes are 19 and 25.

**2)**Other nodes, may be an ancestor of target, or a node in some other subtree. For target node 18 and k is 2, the node 2 comes in this category.

Finding the first type of nodes is easy to implement. Just traverse subtrees rooted with the target node and decrement k in recursive call. When the k becomes 0, print the node currently being traversed (See this for more details). Here we call the function as

*printkdistanceNodeDown()*.

How to find nodes of second type? For the output nodes not lying in the subtree with the target node as the root, we must go through all ancestors. For every ancestor, we find its distance from target node, let the distance be d, now we go to other subtree (if target was found in left subtree, then we go to right subtree and vice versa) of the ancestor and find all nodes at k-d distance from the ancestor.

// Recursive function to print all the nodes at distance k in the // tree (or subtree) rooted with given root. See void printkdistanceNodeDown(Node root, int k) { // Base Case if (root == null || k < 0) return; // If we reach a k distant node, print it if (k==0) { System.out.println(root.data); return; } // Recur for left and right subtrees printkdistanceNodeDown(root.left, k-1); printkdistanceNodeDown(root.right, k-1); } // Prints all nodes at distance k from a given target node. // The k distant nodes may be upward or downward. This function // Returns distance of root from target node, it returns -1 if target // node is not present in tree rooted with root. int printkdistanceNode(Node root, Node target , int k) { // Base Case 1: If tree is empty, return -1 if (root == null) return -1; // If target is same as root. Use the downward function // to print all nodes at distance k in subtree rooted with // target or root if (root == target) { printkdistanceNodeDown(root, k); return 0; } // Recur for left subtree int dl = printkdistanceNode(root.left, target, k); // Check if target node was found in left subtree if (dl != -1) { // If root is at distance k from target, print root // Note that dl is Distance of root's left child from target if (dl + 1 == k) System.out.println(root.data) endl; // Else go to right subtree and print all k-dl-2 distant nodes // Note that the right child is 2 edges away from left child else printkdistanceNodeDown(root.right, k-dl-2); // Add 1 to the distance and return value for parent calls return 1 + dl; } // MIRROR OF ABOVE CODE FOR RIGHT SUBTREE // Note that we reach here only when node was not found in left subtree int dr = printkdistanceNode(root.right, target, k); if (dr != -1) { if (dr + 1 == k) System.out.println(root.data) endl; else printkdistanceNodeDown(root.left, k-dr-2); return 1 + dr; } // If target was neither present in left nor in right subtree return -1; }

**Method 2 - Using the queue**

Use a queue of size K to store the root to node path.

Now since, the queue is of size K.As soon as we find the NODE in tree, the node at front of queue is at a distance K from NODE. It can be the case that the front node is less than K distant from NODE.

So, maintain a counter.

Now start popping a node from queue which is at distant i from NODE, and print all downwards nodes at distance K-i in its other subtree.We only need to print the nodes in other subtree to avoid Error.

Note :- Since we need to print the nodes in sorted order, we can maintain a priority queue to store the nodes and after processing the nodes, we can print it.

**References**

## 0 comments:

## Post a Comment