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
/ \ \
4 5 6
Input: Root of above Binary Tree
k = 4
Output: The tree should be changed to following
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: i) This node becomes a leaf node in which case it needs to be deleted. ii) 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++ program to remove nodes on root to leaf paths of length < K
Node *left, *right;
//New node of a tree
Node *node = newNode;
node->data = data;
node->left = node->right = NULL;
// Utility method that actually removes the nodes which are not
// on the pathLen >= k. This method can change the root as well.
// Method which calls the utitlity method to remove the short path
let pathLen = 0;
returnremoveShortPathNodesUtil(node, 1, k);
//Method to print the tree in inorder fashion.
if(node != null)
document.write(node.data + " ");
let 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);
document.write("The inorder traversal of Original tree is : "+
let res = removeShortPathNodes(root, k);
document.write("The inorder traversal of Modified tree is : "+
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.
This article is contributed by Kumar Gautam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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