Print all nodes at distance k from a given node

#1


getting wrong answer in two testcases.
pls tell the correct code.

#2

You did’nt consider the cases when there will be no such nodes at k distances as in those cases you have to print 0 and also u have to print your answer in sorted order

#3

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#4

pls tell in the form of a code

#5

@yutikakhanna You can have a look at this.
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) 
{ 
    cout << root->data << endl; 
    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) 
        cout << 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) 
        cout << 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;

}