Table of Contents
If you want to practice data structure and algorithm programs, you can go through 100+ Java coding interview questions.
1. Introduction
2. What are Leaf Nodes?
Leaf nodes are the nodes which don’t have any children. These are also terminal nodes and denote the end of the tree structure.
3. Implementation
There are two ways to count leaf nodes in a binary tree:
- Recursive
- Iterative
4.1 Recursive Solution
The recursive solution is very straightforward. Here are the steps for counting the number of leaf nodes:
- If the node is null, then return 0.
- If encountered a leaf node (i.e., node.left is null and node.right is null), then return 1.
- Recursively calculate the number of leaf nodes using
The number of leaf nodes = number of leaf nodes in the left subtree + number of leaf nodes in the right subtree
Code for recursion will be:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 |
// Recursive Solution /* To get the count of leaf nodes in a binary tree */ public static int getLeafCountOfBinaryTree(TreeNode node) { if(node == null) return 0; if(node.left == null && node.right == null) return 1; else return getLeafCountOfBinaryTree(node.left) + getLeafCountOfBinaryTree(node.right); } |
The space complexity is O(w) where w is the maximum width in the binary tree. This is because the maximum number of nodes in the stack is proportional to the width of the binary tree at any point in time.
4.2 Iterative Solution
For recursion, we use an implicit stack. So here, to convert the recursive solution to iterative, we will use an explicit stack.
Steps for the iterative solution:
- Create an empty
stackand push the root node to it. - Initialize a variable
countto track the number of leaf nodes. - Do the following when the
stackis not empty:- Pop a node from the
stackand check if it is a leaf node. If yes, increase the count. - Push the
right childof the popped node to thestack. - Push the
left childof the popped node to the stack.
- Pop a node from the
We are pushing the right child first, so it will be processed after the left subtree as the Stack is LIFO (Last In, First Out).
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
int countLeavesIterative(TreeNode root) { if (root == null) return 0; Stack<TreeNode> stack = new Stack<>(); stack.push(root); int count = 0; while (!stack.isEmpty()) { TreeNode current = stack.pop(); if (current.left == null && current.right == null) { count++; } if (current.right != null) { stack.push(current.right); } if (current.left != null) { stack.push(current.left); } } return count; } |
The time complexity for iterative approach is O(n), where n is the number of nodes in the binary tree. This is because each node in the tree is visited to determine whether it is a leaf node or not.
The space complexity is O(w) where w is the maximum width in the binary tree. This is because the maximum number of nodes in the stack is proportional to the width of the binary tree at any point in time.
5. Complete Java Program
Let’s say our binary tree is:
Here is complete java program for counting number of leaf nodes:
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 |
package org.arpit.java2blog.generic; import java.util.Stack; public class BinaryTreeLeafCount { public static class TreeNode { int data; TreeNode left; TreeNode right; TreeNode(int data) { this.data=data; } } // Recursive Solution /* To get the count of leaf nodes in a binary tree*/ public static int getLeafCountOfBinaryTree(TreeNode node) { if(node == null) return 0; if(node.left ==null && node.right==null) return 1; else return getLeafCountOfBinaryTree(node.left)+ getLeafCountOfBinaryTree(node.right); } // Iterative solution public static int getLeafCountOfBinaryTreeIterative(TreeNode root) { if (root == null) return 0; Stack<TreeNode> stack = new Stack<>(); stack.push(root); int count = 0; while (!stack.isEmpty()) { TreeNode current = stack.pop(); if (current.left == null && current.right == null) { count++; } if (current.right != null) { stack.push(current.right); } if (current.left != null) { stack.push(current.left); } } return count; } public static void main(String[] args) { // Creating a binary tree TreeNode rootNode=createBinaryTree(); System.out.println("Number of leaf nodes in binary tree(Recursive) :"+getLeafCountOfBinaryTree(rootNode)); System.out.println("Number of leaf nodes in binary tree(Iterative) :"+getLeafCountOfBinaryTreeIterative(rootNode)); } public static TreeNode createBinaryTree() { TreeNode rootNode =new TreeNode(40); TreeNode node20=new TreeNode(20); TreeNode node10=new TreeNode(10); TreeNode node30=new TreeNode(30); TreeNode node60=new TreeNode(60); TreeNode node50=new TreeNode(50); TreeNode node70=new TreeNode(70); rootNode.left=node20; rootNode.right=node60; node20.left=node10; node20.right=node30; node60.left=node50; node60.right=node70; return rootNode; } } |
Run above program and you will get following output:
|
1 2 3 4 |
Number of leaf nodes in binary tree(Recursive) :4 Number of leaf nodes in binary tree(Iterative) :4 |
6. Conclusion
In this article, we covered about how to count leaf nodes in Binary tree. We have discussed two approaches: Iterative and Recursive. We also discussed about time and space complexity for the same.
Please go through Frequently asked java interview programs  for more such programs.
