Given an N-ary tree T of N nodes, the task is to calculate the longest path between any two nodes(also known as the diameter of the tree).
Different approaches to solve these problems have already been discussed:
In this post, we will be discussing an approach which uses Dynamic Programming on Trees.
There are two possibilities for the diameter to exist:
- Case 1: Suppose the diameter starts from a node and ends at some node in its subtree. Let’s say that there exist a node x such that the longest path starts from node x and goes into its subtree and ends at some node in the subtree itself. Let’s define this path length by dp1[x].
- Case 2: Suppose the diameter or the longest path starts in subtree of a node x, passes through it and ends in it’s subtree. Let’s define this path by dp2[x].
If for all nodes x, we take a maximum of dp1[x], dp2[x], then we will get the diameter of the tree.
For the case-1, to find dp1[node], we need to find the maximum of all dp1[x], where x is the children of node. And dp1[node] will be equal to 1 + max(dp1[children1], dp1[children2], ..).
For the case-2, to find dp2[node], we need to find the two maximum of all dp1[x], where x is the children of node. And dp2[node] will be equal to 1 + max 2 of(dp1[children1], dp1[children2], ..).
We can easily run a DFS and find the maximum of both dp1[node] and dp2[node] for every to get the diameter of the tree.
Below is the implementation of the above approach:
Diameter of the given tree is 4
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- Diameter of a tree using DFS
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- Generic Trees(N-array Trees)
- Total number of possible Binary Search Trees and Binary Trees with n keys
- Maximum sub-tree sum in a Binary Tree such that the sub-tree is also a BST
- Complexity of different operations in Binary tree, Binary Search Tree and AVL tree
- Print Binary Tree levels in sorted order | Set 3 (Tree given as array)
- Convert an arbitrary Binary Tree to a tree that holds Children Sum Property
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Improved By : rituraj_jain