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.
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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 :
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.