Given a binary tree, we need to write a program to print all leaf nodes of the given binary tree from left to right. That is, the nodes should be printed in the order they appear from left to right in the given tree.
For Example,
For the above binary tree, the output will be as shown below:
4 6 7 9 10
The idea to do this is similar to DFS algorithm. Below is a step by step algorithm to do this:
- Check if the given node is null. If null, then return from the function.
- Check if it is a leaf node. If the node is a leaf node, then print its data.
- If in the above step, the node is not a leaf node then check if the left and right children of node exist. If yes then call the function for left and right child of the node recursively.
Below is the implementation of the above approach.
C++
/* C++ program to print leaf nodes from left to right */#include <iostream> using namespace std; // A Binary Tree Node struct Node { int data; struct Node *left, *right; }; // function to print leaf // nodes from left to right void printLeafNodes(Node *root) { // if node is null, return if (!root) return; // if node is leaf node, print its data if (!root->left && !root->right) { cout << root->data << " "; return; } // if left child exists, check for leaf // recursively if (root->left) printLeafNodes(root->left); // if right child exists, check for leaf // recursively if (root->right) printLeafNodes(root->right); } // Utility function to create a new tree node Node* newNode(int data) { Node *temp = new Node; temp->data = data; temp->left = temp->right = NULL; return temp; } // Driver program to test above functions int main() { // Let us create binary tree shown in // above diagram Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->right->left = newNode(5); root->right->right = newNode(8); root->right->left->left = newNode(6); root->right->left->right = newNode(7); root->right->right->left = newNode(9); root->right->right->right = newNode(10); // print leaf nodes of the given tree printLeafNodes(root); return 0; } |
Java
// Java program to print leaf nodes // from left to right import java.util.*; class GFG{ // A Binary Tree Node static class Node { public int data; public Node left, right; }; // Function to print leaf // nodes from left to right static void printLeafNodes(Node root) { // If node is null, return if (root == null) return; // If node is leaf node, print its data if (root.left == null && root.right == null) { System.out.print(root.data + " "); return; } // If left child exists, check for leaf // recursively if (root.left != null) printLeafNodes(root.left); // If right child exists, check for leaf // recursively if (root.right != null) printLeafNodes(root.right); } // Utility function to create a new tree node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Driver code public static void main(String []args) { // Let us create binary tree shown in // above diagram Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.right.left = newNode(5); root.right.right = newNode(8); root.right.left.left = newNode(6); root.right.left.right = newNode(7); root.right.right.left = newNode(9); root.right.right.right = newNode(10); // Print leaf nodes of the given tree printLeafNodes(root); } } // This code is contributed by pratham76 |
Python3
# Python3 program to print # leaf nodes from left to right # Binary tree node class Node: def __init__(self, data): self.data = data self.left = None self.right = None # Function to print leaf # nodes from left to right def printLeafNodes(root: Node) -> None: # If node is null, return if (not root): return # If node is leaf node, # print its data if (not root.left and not root.right): print(root.data, end = " ") return # If left child exists, # check for leaf recursively if root.left: printLeafNodes(root.left) # If right child exists, # check for leaf recursively if root.right: printLeafNodes(root.right) # Driver Code if __name__ == "__main__": # Let us create binary tree shown in # above diagram root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.right.left = Node(5) root.right.right = Node(8) root.right.left.left = Node(6) root.right.left.right = Node(7) root.right.right.left = Node(9) root.right.right.right = Node(10) # print leaf nodes of the given tree printLeafNodes(root) # This code is contributed by sanjeev2552 |
C#
// C# program to print leaf nodes // from left to right using System; class GFG{ // A Binary Tree Node class Node { public int data; public Node left, right; }; // Function to print leaf // nodes from left to right static void printLeafNodes(Node root) { // If node is null, return if (root == null) return; // If node is leaf node, print its data if (root.left == null && root.right == null) { Console.Write(root.data + " "); return; } // If left child exists, check for leaf // recursively if (root.left != null) printLeafNodes(root.left); // If right child exists, check for leaf // recursively if (root.right != null) printLeafNodes(root.right); } // Utility function to create a new tree node static Node newNode(int data) { Node temp = new Node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Driver code public static void Main() { // Let us create binary tree shown in // above diagram Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.right.left = newNode(5); root.right.right = newNode(8); root.right.left.left = newNode(6); root.right.left.right = newNode(7); root.right.right.left = newNode(9); root.right.right.right = newNode(10); // Print leaf nodes of the given tree printLeafNodes(root); } } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript program to print leaf nodes // from left to right // A Binary Tree Node class Node { constructor() { this.data = 0; this.left = null; this.right = null; } }; // Function to print leaf // nodes from left to right function printLeafNodes(root) { // If node is null, return if (root == null) return; // If node is leaf node, print its data if (root.left == null && root.right == null) { document.write(root.data + " "); return; } // If left child exists, check for leaf // recursively if (root.left != null) printLeafNodes(root.left); // If right child exists, check for leaf // recursively if (root.right != null) printLeafNodes(root.right); } // Utility function to create a new tree node function newNode(data) { var temp = new Node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Driver code // Let us create binary tree shown in // above diagram var root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.right.left = newNode(5); root.right.right = newNode(8); root.right.left.left = newNode(6); root.right.left.right = newNode(7); root.right.right.left = newNode(9); root.right.right.right = newNode(10); // Print leaf nodes of the given tree printLeafNodes(root); // This code is contributed by rrrtnx </script> |
Output
4 6 7 9 10
Time Complexity: O(n), where n is the number of nodes in the binary tree.
Auxiliary Space: O(n)
Iterative Method:
Following is a simple stack-based iterative method to print the leaf nodes from left to right.
- Create an empty stack ‘st’ and push the root node to stack.
- Do the following while stack is not empty.
- Pop an item from the stack
- If the node is a leaf node then print it.
- Else:
- If the right node is not NULL
- push the right node into the stack
- If the left node is not NULL
- push the left node into the stack
- If the right node is not NULL
Below is the implementation of the above approach.
C++
// C++ program to print leaf nodes from left to right #include <bits/stdc++.h> using namespace std; // A Binary Tree Node struct Node { int data; struct Node *left, *right; }; // fun to creates and returns a new node Node* newNode(int data) { Node* temp = new Node(); temp->data = data; temp->left = temp->right = NULL; return temp; } // fun to print leaf nodes from left to right void printleafNodes(Node* root) { // base case if (!root) return; // implement iterative preorder traversal and start // storing leaf nodes. stack<Node*> st; st.push(root); while (!st.empty()) { root = st.top(); st.pop(); // if node is leafnode, print its data if (!root->left && !root->right) cout << root->data << " "; if (root->right) st.push(root->right); if (root->left) st.push(root->left); } } // Driver program to test above functions int main() { // create a binary tree Node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->right->left = newNode(5); root->right->right = newNode(8); root->right->left->left = newNode(6); root->right->left->right = newNode(7); root->right->right->left = newNode(9); root->right->right->right = newNode(10); // prints leaf nodes of the given tree printleafNodes(root); return 0; } // This Method and code is contributed by Modem Upendra |
Java
// Java program to print leaf nodes from left to right import java.util.*; public class GFG { // structure of node of Binary Tree static class Node { int data; Node left, right; Node(int item) { data = item; left = right = null; } } // fun to print leaf nodes from left to right static void printleafNodes(Node root) { // base case if (root == null) return; // implement iterative preorder traversal and start // storing leaf nodes. Stack<Node> st = new Stack<>(); st.push(root); while (!st.isEmpty()) { root = st.peek(); st.pop(); // if node is leafnode, print its data if (root.left == null && root.right == null) System.out.print(root.data + " "); if (root.right != null) st.push(root.right); if (root.left != null) st.push(root.left); } } // Driver program to test above functions public static void main(String[] args) { // create a binary tree Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.right.left = new Node(5); root.right.right = new Node(8); root.right.left.left = new Node(6); root.right.left.right = new Node(7); root.right.right.left = new Node(9); root.right.right.right = new Node(10); // prints leaf nodes of the given tree printleafNodes(root); } } // This code is contributed by Karandeep1234 |
Python3
# Python3 program to print leaf nodes from left to right # A Binary Tree Node class Node: def __init__(self, data): self.data = data self.left = None self.right = None # fun to creates and returns a new node def newNode(data): temp = Node(data) return temp # fun to print leaf nodes from left to right def printleafNodes(root): # base case if not root: return # implement iterative preorder traversal and start # storing leaf nodes. st = [] st.append(root) while len(st) > 0: root = st.pop() # if node is leafnode, print its data if not root.left and not root.right: print(root.data, end=" ") if root.right: st.append(root.right) if root.left: st.append(root.left) # Driver program to test above functions if __name__ == '__main__': # create a binary tree root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.right.left = newNode(5) root.right.right = newNode(8) root.right.left.left = newNode(6) root.right.left.right = newNode(7) root.right.right.left = newNode(9) root.right.right.right = newNode(10) # prints leaf nodes of the given tree printleafNodes(root) # This code is contributed by lokeshpotta20. |
C#
using System; using System.Collections.Generic; class GFG { // structure of node of Binary Tree class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; } } // fun to print leaf nodes from left to right static void printleafNodes(Node root) { // base case if (root == null) return; // implement iterative preorder traversal and start // storing leaf nodes. Stack<Node> st = new Stack<Node>(); st.Push(root); while (st.Count != 0) { root = st.Peek(); st.Pop(); // if node is leafnode, print its data if (root.left == null && root.right == null) Console.Write(root.data + " "); if (root.right != null) st.Push(root.right); if (root.left != null) st.Push(root.left); } } // Driver program to test above functions public static void Main(string[] args) { // create a binary tree Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.right.left = new Node(5); root.right.right = new Node(8); root.right.left.left = new Node(6); root.right.left.right = new Node(7); root.right.right.left = new Node(9); root.right.right.right = new Node(10); // prints leaf nodes of the given tree printleafNodes(root); } } // This code is contributed by pradeepkumarppk2003 |
Javascript
// Javascript program to print leaf nodes from left to right // A Binary Tree Node class Node { constructor(data){ this.data=data; this.left=null; this.right=null; } } // fun to print leaf nodes from left to right function printleafNodes(root) { // base case if (!root) return; // implement iterative preorder traversal and start // storing leaf nodes. let st=[]; st.push(root); while (st.length>0) { root = st[st.length-1]; st.pop(); // if node is leafnode, print its data if (!root.left && !root.right) document.write(root.data + " "); if (root.right) st.push(root.right); if (root.left) st.push(root.left); } } // Driver program to test above functions // create a binary tree let root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.right.left = new Node(5); root.right.right = new Node(8); root.right.left.left = new Node(6); root.right.left.right = new Node(7); root.right.right.left = new Node(9); root.right.right.right = new Node(10); // prints leaf nodes of the given tree printleafNodes(root); |
Output
4 6 7 9 10
Time Complexity: O(n), where n is the number of nodes in the binary tree.
Auxiliary Space: O(n)
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