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


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, output will be as shown below:

4 6 7 9 10

The idea to do this is similar to DFS algorithm. Below is step by step algorithm to do this:



  1. Check if given node is null. If null, then return from the function.
  2. Check if it is a leaf node. If the node is a leaf node, then print its data.
  3. If in above step, node is not a leaf node then check if left and right childs of node exists. If yes then call function for left and right childs of the node recursively.

Below is C++ implementation of above approach.

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/* 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;
}

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Output:

4 6 7 9 10

Time Complexity: O( n ) , where n is number of nodes in the binary tree.

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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