Print all leaf nodes of a Binary Tree from left to right - GeeksforGeeks

29 Jun, 2021
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.
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++
Java
Python3
C#
Javascript
/* C++ program to print leaf nodes from left   to right */#include <iostream>using namespace std;  // A Binary Tree Nodestruct Node{    int data;    struct Node *left, *right;}; // function to print leaf// nodes from left to rightvoid 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 nodeNode* newNode(int data){    Node *temp = new Node;    temp->data = data;    temp->left = temp->right = NULL;    return temp;}  // Driver program to test above functionsint 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 program to print leaf nodes// from left to rightimport java.util.*;   class GFG{      // A Binary Tree Nodestatic class Node{    public int data;    public Node left, right;};  // Function to print leaf// nodes from left to rightstatic 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 nodestatic Node newNode(int data){    Node temp = new Node();    temp.data = data;    temp.left = null;    temp.right = null;    return temp;}  // Driver codepublic 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 program to print# leaf nodes from left to right # Binary tree nodeclass Node:       def __init__(self, data):        self.data = data        self.left = None        self.right = None # Function to print leaf# nodes from left to rightdef 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 Codeif __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# program to print leaf nodes// from left to rightusing System;  class GFG{     // A Binary Tree Nodeclass Node{    public int data;    public Node left, right;}; // Function to print leaf// nodes from left to rightstatic 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 nodestatic Node newNode(int data){    Node temp = new Node();    temp.data = data;    temp.left = null;    temp.right = null;    return temp;} // Driver codepublic 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
<script> // Javascript program to print leaf nodes// from left to right // A Binary Tree Nodeclass Node{    constructor()    {        this.data = 0;        this.left = null;        this.right = null;    }}; // Function to print leaf// nodes from left to rightfunction 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 nodefunction 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 diagramvar 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 treeprintLeafNodes(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. 
https://youtu.be/rBekvLt23wY?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbkThis article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
 
sanjeev2552
rutvik_56
pratham76
rrrtnx
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