Given an N-ary tree, print the number of leaf nodes in the subtree of every node.
Input: 1 / \ 2 3 / | \ 4 5 6 Output: The node 1 has 4 leaf nodes The node 2 has 1 leaf nodes The node 3 has 3 leaf nodes The node 4 has 1 leaf nodes The node 5 has 1 leaf nodes The node 6 has 1 leaf nodes
Approach: The idea is to perform DFS traversal on the given tree and for every node keep an array leaf to store the count of leaf nodes in the subtree below it.
Now, while recurring down the tree, if a leaf node is found set it’s leaf[i] value to 1 and return back in upward direction. Now, every time while returning back from the function call to upward, add the leaf nodes of the node below it.
Once, the DFS traversal is completed we will have the count of leaf nodes in the array leaf.
Below is the implementation of the above approach:
The node 1 has 4 leaf nodes The node 2 has 1 leaf nodes The node 3 has 3 leaf nodes The node 4 has 1 leaf nodes The node 5 has 1 leaf nodes The node 6 has 1 leaf nodes
- Convert a Binary Tree such that every node stores the sum of all nodes in its right subtree
- Change a Binary Tree so that every node stores sum of all nodes in left subtree
- Print the nodes of binary tree as they become the leaf node
- Check if two nodes are in same subtree of the root node
- Subtree of all nodes in a tree using DFS
- Print all nodes that are at distance k from a leaf node
- Sum of nodes on the longest path from root to leaf node
- Closest leaf to a given node in Binary Tree
- Deepest left leaf node in a binary tree
- Sum of all leaf nodes of binary tree
- Deepest right leaf node in a binary tree | Iterative approach
- Count Non-Leaf nodes in a Binary Tree
- Determine the count of Leaf nodes in an N-ary tree
- Product of all leaf nodes of binary tree
- Deepest left leaf node in a binary tree | iterative approach
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