Given the value of ‘N’ and ‘I’. Here, represents the number of internal nodes present in an N-ary tree and every node of the N-ary can either have childs or zero child. The task is to determine the number of Leaf nodes in n-ary tree.
Input : N = 3, I = 5
Output : Leaf nodes = 11
Input : N = 10, I = 10
Output : leaf nodes = 91
I = Number of Internal nodes.
L = Leaf Nodes.
and, N = Number of children each node can have.
Derivation: The tree is an N-ary tree. Assume it has T total nodes, which is the sum of internal nodes (I) and leaf nodes (L). A tree with T total nodes will have (T – 1) edges or branches.
In other words, since the tree is an N-ary tree, each internal node will have N branches contributing a total of N*I internal branches. Therefore we have the following relations from the above explanations,
- N * I = T – 1
- L + I = T
From the above two equations, we can say that L = (N – 1) * I + 1.
Below is the implementation of the above approach:
Leaf nodes = 9
- Count Non-Leaf nodes in a Binary Tree
- Program to count leaf nodes in a binary tree
- Iterative program to count leaf nodes in a Binary Tree
- Sum of all leaf nodes of binary tree
- Print all leaf nodes of an n-ary tree using DFS
- Product of all leaf nodes of binary tree
- Print all leaf nodes of a binary tree from right to left
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Maximum sum of leaf nodes among all levels of the given binary tree
- Pairwise Swap leaf nodes in a binary tree
- Print the nodes of binary tree as they become the leaf node
- Maximum sum of non-leaf nodes among all levels of the given binary tree
- Print all leaf nodes of a Binary Tree from left to right
- Print Sum and Product of all Non-Leaf nodes in Binary Tree
- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
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