Given a Binary Tree and a number k, remove all nodes that lie only on root to leaf path(s) of length smaller than k. If a node X lies on multiple root-to-leaf paths and if any of the paths has path length >= k, then X is not deleted from Binary Tree. In other words a node is deleted if all paths going through it have lengths smaller than k.
Consider the following example Binary Tree
1
/ \
2 3
/ \ \
4 5 6
/ /
7 8
Input: Root of above Binary Tree
k = 4
Output: The tree should be changed to following
1
/ \
2 3
/ \
4 6
/ /
7 8
There are 3 paths
i) 1->2->4->7 path length = 4
ii) 1->2->5 path length = 3
iii) 1->3->6->8 path length = 4
There is only one path " 1->2->5 " of length smaller than 4.
The node 5 is the only node that lies only on this path, so
node 5 is removed.
Nodes 2 and 1 are not removed as they are parts of other paths
of length 4 as well.
If k is 5 or greater than 5, then whole tree is deleted.
If k is 3 or less than 3, then nothing is deleted.We strongly recommend to minimize your browser and try this yourself first
The idea here is to use post order traversal of the tree. Before removing a node we need to check that all the children of that node in the shorter path are already removed.
There are 2 cases:
- This node becomes a leaf node in which case it needs to be deleted.
- This node has other child on a path with path length >= k. In that case it needs not to be deleted
The implementation of above approach is as below :
C++
// C++ program to remove nodes on root to leaf paths of length < K#include<bits/stdc++.h>using namespace std;
struct Node
{ int data;
Node *left, *right;
};//New node of a treeNode *newNode(int data)
{ Node *node = new Node;
node->data = data;
node->left = node->right = NULL;
return node;
}// Utility method that actually removes the nodes which are not// on the pathLen >= k. This method can change the root as well.Node *removeShortPathNodesUtil(Node *root, int level, int k)
{ //Base condition
if (root == NULL)
return NULL;
// Traverse the tree in postorder fashion so that if a leaf
// node path length is shorter than k, then that node and
// all of its descendants till the node which are not
// on some other path are removed.
root->left = removeShortPathNodesUtil(root->left, level + 1, k);
root->right = removeShortPathNodesUtil(root->right, level + 1, k);
// If root is a leaf node and it's level is less than k then
// remove this node.
// This goes up and check for the ancestor nodes also for the
// same condition till it finds a node which is a part of other
// path(s) too.
if (root->left == NULL && root->right == NULL && level < k)
{
delete root;
return NULL;
}
// Return root;
return root;
}// Method which calls the utility method to remove the short path// nodes.Node *removeShortPathNodes(Node *root, int k)
{ int pathLen = 0;
return removeShortPathNodesUtil(root, 1, k);
}//Method to print the tree in inorder fashion.void printInorder(Node *root)
{ if (root)
{
printInorder(root->left);
cout << root->data << " ";
printInorder(root->right);
}
}// Driver method.int main()
{ int k = 4;
Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->left->left->left = newNode(7);
root->right->right = newNode(6);
root->right->right->left = newNode(8);
cout << "Inorder Traversal of Original tree" << endl;
printInorder(root);
cout << endl;
cout << "Inorder Traversal of Modified tree" << endl;
Node *res = removeShortPathNodes(root, k);
printInorder(res);
return 0;
} |
Java
// Java program to remove nodes on root to leaf paths of length < k /* Class containing left and right child of current node and key value*/
class Node
{ int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
} class BinaryTree
{ Node root;
// Utility method that actually removes the nodes which are not
// on the pathLen >= k. This method can change the root as well.
Node removeShortPathNodesUtil(Node node, int level, int k)
{
//Base condition
if (node == null)
return null;
// Traverse the tree in postorder fashion so that if a leaf
// node path length is shorter than k, then that node and
// all of its descendants till the node which are not
// on some other path are removed.
node.left = removeShortPathNodesUtil(node.left, level + 1, k);
node.right = removeShortPathNodesUtil(node.right, level + 1, k);
// If root is a leaf node and it's level is less than k then
// remove this node.
// This goes up and check for the ancestor nodes also for the
// same condition till it finds a node which is a part of other
// path(s) too.
if (node.left == null && node.right == null && level < k)
return null;
// Return root;
return node;
}
// Method which calls the utility method to remove the short path
// nodes.
Node removeShortPathNodes(Node node, int k)
{
int pathLen = 0;
return removeShortPathNodesUtil(node, 1, k);
}
//Method to print the tree in inorder fashion.
void printInorder(Node node)
{
if (node != null)
{
printInorder(node.left);
System.out.print(node.data + " ");
printInorder(node.right);
}
}
// Driver program to test for samples
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
int k = 4;
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.left.left.left = new Node(7);
tree.root.right.right = new Node(6);
tree.root.right.right.left = new Node(8);
System.out.println("The inorder traversal of original tree is : ");
tree.printInorder(tree.root);
Node res = tree.removeShortPathNodes(tree.root, k);
System.out.println("");
System.out.println("The inorder traversal of modified tree is : ");
tree.printInorder(res);
}
} // This code has been contributed by Mayank Jaiswal |
Python3
# Python3 program to remove nodes on root # to leaf paths of length < K # New node of a tree class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# Utility method that actually removes # the nodes which are not on the pathLen >= k.# This method can change the root as well. def removeShortPathNodesUtil(root, level, k) :
# Base condition
if (root == None) :
return None
# Traverse the tree in postorder fashion
# so that if a leaf node path length is
# shorter than k, then that node and all
# of its descendants till the node which
# are not on some other path are removed.
root.left = removeShortPathNodesUtil(root.left,
level + 1, k)
root.right = removeShortPathNodesUtil(root.right,
level + 1, k)
# If root is a leaf node and it's level
# is less than k then remove this node.
# This goes up and check for the ancestor
# nodes also for the same condition till
# it finds a node which is a part of other
# path(s) too.
if (root.left == None and
root.right == None and level < k) :
return None
# Return root
return root
# Method which calls the utility method # to remove the short path nodes. def removeShortPathNodes(root, k) :
pathLen = 0
return removeShortPathNodesUtil(root, 1, k)
# Method to print the tree in# inorder fashion. def printInorder(root) :
if (root) :
printInorder(root.left)
print(root.data, end = " " )
printInorder(root.right)
# Driver Code if __name__ == '__main__':
k = 4
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)
root.left.left.left = newNode(7)
root.right.right = newNode(6)
root.right.right.left = newNode(8)
print("Inorder Traversal of Original tree" )
printInorder(root)
print()
print("Inorder Traversal of Modified tree" )
res = removeShortPathNodes(root, k)
printInorder(res)
# This code is contributed # by SHUBHAMSINGH10 |
C#
using System;
// C# program to remove nodes on root to leaf paths of length < k /* Class containing left and right child of current node and key value*/
public class Node
{ public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}public class BinaryTree
{ public Node root;
// Utility method that actually removes the nodes which are not
// on the pathLen >= k. This method can change the root as well.
public virtual Node removeShortPathNodesUtil(Node node, int level, int k)
{
//Base condition
if (node == null)
{
return null;
}
// Traverse the tree in postorder fashion so that if a leaf
// node path length is shorter than k, then that node and
// all of its descendants till the node which are not
// on some other path are removed.
node.left = removeShortPathNodesUtil(node.left, level + 1, k);
node.right = removeShortPathNodesUtil(node.right, level + 1, k);
// If root is a leaf node and it's level is less than k then
// remove this node.
// This goes up and check for the ancestor nodes also for the
// same condition till it finds a node which is a part of other
// path(s) too.
if (node.left == null && node.right == null && level < k)
{
return null;
}
// Return root;
return node;
}
// Method which calls the utility method to remove the short path
// nodes.
public virtual Node removeShortPathNodes(Node node, int k)
{
int pathLen = 0;
return removeShortPathNodesUtil(node, 1, k);
}
//Method to print the tree in inorder fashion.
public virtual void printInorder(Node node)
{
if (node != null)
{
printInorder(node.left);
Console.Write(node.data + " ");
printInorder(node.right);
}
}
// Driver program to test for samples
public static void Main(string[] args)
{
BinaryTree tree = new BinaryTree();
int k = 4;
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.root.left.left.left = new Node(7);
tree.root.right.right = new Node(6);
tree.root.right.right.left = new Node(8);
Console.WriteLine("The inorder traversal of original tree is : ");
tree.printInorder(tree.root);
Node res = tree.removeShortPathNodes(tree.root, k);
Console.WriteLine("");
Console.WriteLine("The inorder traversal of modified tree is : ");
tree.printInorder(res);
}
} // This code is contributed by Shrikant13
|
Javascript
<script> // JavaScript program to remove nodes on
// root to leaf paths of length < k
class Node
{
constructor(item) {
this.left = null;
this.right = null;
this.data = item;
}
}
let root;
// Utility method that actually removes the nodes which are not
// on the pathLen >= k. This method can change the root as well.
function removeShortPathNodesUtil(node, level, k)
{
//Base condition
if (node == null)
return null;
// Traverse the tree in postorder fashion so that if a leaf
// node path length is shorter than k, then that node and
// all of its descendants till the node which are not
// on some other path are removed.
node.left = removeShortPathNodesUtil(node.left, level + 1, k);
node.right = removeShortPathNodesUtil(node.right, level + 1, k);
// If root is a leaf node and it's level is less than k then
// remove this node.
// This goes up and check for the ancestor nodes also for the
// same condition till it finds a node which is a part of other
// path(s) too.
if (node.left == null && node.right == null && level < k)
return null;
// Return root;
return node;
}
// Method which calls the utility method to remove the short path
// nodes.
function removeShortPathNodes(node, k)
{
let pathLen = 0;
return removeShortPathNodesUtil(node, 1, k);
}
//Method to print the tree in inorder fashion.
function printInorder(node)
{
if (node != null)
{
printInorder(node.left);
document.write(node.data + " ");
printInorder(node.right);
}
}
let k = 4;
root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.left.left.left = new Node(7);
root.right.right = new Node(6);
root.right.right.left = new Node(8);
document.write("The inorder traversal of Original tree is : " +
"</br>");
printInorder(root);
let res = removeShortPathNodes(root, k);
document.write("</br>");
document.write("The inorder traversal of Modified tree is : " +
"</br>");
printInorder(res);
</script> |
Output
Inorder Traversal of Original tree 7 4 2 5 1 3 8 6 Inorder Traversal of Modified tree 7 4 2 1 3 8 6
Time complexity of the above solution is O(n) where n is number of nodes in given Binary Tree.
Auxiliary Space: If we don’t consider the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree.