Given a Binary tree, print all the leaf nodes of a Binary tree at a given level L.
Examples:
Input:
1
/ \
2 3
/ / \
4 5 6
level = 3
Output: 4 5 6
Input:
7
/ \
2 3
/ \ \
4 9 10
/
6
level = 3
Output: 4 9
Approach: Recursively traverse the tree in a level order manner. If the current level is the same as the given level, then check whether the current node is a leaf node or not. If it is a leaf node then print it.
Below is the implementation of the above approach:Â
C++
// C++ program to print all the// leaf nodes at a given level// in a Binary tree#include <bits/stdc++.h>using namespace std;Â
// Binary tree nodestruct node {Â Â Â Â struct node* left;Â Â Â Â struct node* right;Â Â Â Â int data;};Â
// Function to create a new// Binary nodestruct node* newNode(int data){Â Â Â Â struct node* temp = new node;Â
    temp->data = data;    temp->left = NULL;    temp->right = NULL;Â
    return temp;}Â
// Function to print Leaf Nodes at// a given levelvoid PrintLeafNodes(struct node* root, int level){Â Â Â Â if (root == NULL)Â Â Â Â Â Â Â Â return;Â
    if (level == 1) {        if (root->left == NULL && root->right == NULL)            cout << root->data << " ";    }    else if (level > 1) {        PrintLeafNodes(root->left, level - 1);        PrintLeafNodes(root->right, level - 1);    }}Â
// Driver codeint main(){Â Â Â Â struct node* root = newNode(1);Â Â Â Â root->left = newNode(2);Â Â Â Â root->right = newNode(3);Â Â Â Â root->left->left = newNode(6);Â Â Â Â root->right->right = newNode(4);Â Â Â Â root->left->left->left = newNode(8);Â Â Â Â root->left->left->right = newNode(7);Â
    int level = 4;Â
    PrintLeafNodes(root, level);Â
    return 0;} |
Java
// Java program to print all the // leaf nodes at a given level // in a Binary tree class GFG{Â
    // Binary tree node     static class node    {Â
        node left;        node right;        int data;    };Â
    // Function to create a new     // Binary node     static node newNode(int data)     {        node temp = new node();Â
        temp.data = data;        temp.left = null;        temp.right = null;Â
        return temp;    }Â
    // Function to print Leaf Nodes at     // a given level     static void PrintLeafNodes(node root, int level)     {        if (root == null)        {            return;        }Â
        if (level == 1)         {            if (root.left == null && root.right == null)             {                System.out.print(root.data + " ");            }Â
        }         else if (level > 1)         {            PrintLeafNodes(root.left, level - 1);            PrintLeafNodes(root.right, level - 1);        }    }Â
    // Driver code     public static void main(String[] args)     {        node root = newNode(1);        root.left = newNode(2);        root.right = newNode(3);        root.left.left = newNode(6);        root.right.right = newNode(4);        root.left.left.left = newNode(8);        root.left.left.right = newNode(7);Â
        int level = 4;Â
        PrintLeafNodes(root, level);    }}Â
// This code contributed by Rajput-Ji |
Python3
# Python3 program to print all the# leaf nodes at a given level# in a Binary treeÂ
# Binary tree nodeclass node:         def __init__(self, data):        self.left=None        self.right=None        self.data=data  # Function to create a new# Binary nodedef newNode(data):    return node(data)  # Function to print Leaf Nodes at# a given leveldef PrintLeafNodes(root, level):    if (root == None):        return      if (level == 1):        if (root.left == None and            root.right == None):            print(root.data, end = " ")         elif (level > 1):        PrintLeafNodes(root.left, level - 1)        PrintLeafNodes(root.right, level - 1)         if __name__=="__main__":         root=newNode(1)    root.left = newNode(2);    root.right = newNode(3);    root.left.left = newNode(6);    root.right.right = newNode(4);    root.left.left.left = newNode(8);    root.left.left.right = newNode(7);    level = 4;    PrintLeafNodes(root, level);Â
# This code is contributed by rutvik_56 |
C#
// C# program to print all the // leaf nodes at a given level // in a Binary tree using System;Â
class GFG{Â
    // Binary tree node     public class node    {Â
        public node left;        public node right;        public int data;    };Â
    // Function to create a new     // Binary node     static node newNode(int data)     {        node temp = new node();Â
        temp.data = data;        temp.left = null;        temp.right = null;Â
        return temp;    }Â
    // Function to print Leaf Nodes at     // a given level     static void PrintLeafNodes(node root, int level)     {        if (root == null)        {            return;        }Â
        if (level == 1)         {            if (root.left == null && root.right == null)             {                Console.Write(root.data + " ");            }Â
        }         else if (level > 1)         {            PrintLeafNodes(root.left, level - 1);            PrintLeafNodes(root.right, level - 1);        }    }Â
    // Driver code     public static void Main(String[] args)     {        node root = newNode(1);        root.left = newNode(2);        root.right = newNode(3);        root.left.left = newNode(6);        root.right.right = newNode(4);        root.left.left.left = newNode(8);        root.left.left.right = newNode(7);Â
        int level = 4;Â
        PrintLeafNodes(root, level);    }}Â
// This code has been contributed by 29AjayKumar |
Javascript
<script>Â
    // JavaScript program to print all the    // leaf nodes at a given level    // in a Binary tree         // Binary tree node     class node    {        constructor(data) {           this.left = null;           this.right = null;           this.data = data;        }    }         // Function to create a new    // Binary node    function newNode(data)    {        let temp = new node(data);          temp.data = data;        temp.left = null;        temp.right = null;          return temp;    }      // Function to print Leaf Nodes at    // a given level    function PrintLeafNodes(root, level)    {        if (root == null)        {            return;        }          if (level == 1)        {            if (root.left == null && root.right == null)            {                document.write(root.data + " ");            }          }        else if (level > 1)        {            PrintLeafNodes(root.left, level - 1);            PrintLeafNodes(root.right, level - 1);        }    }         let root = newNode(1);    root.left = newNode(2);    root.right = newNode(3);    root.left.left = newNode(6);    root.right.right = newNode(4);    root.left.left.left = newNode(8);    root.left.left.right = newNode(7);Â
    let level = 4;Â
    PrintLeafNodes(root, level);Â
</script> |
Output
8 7
Complexity Analysis:
- Time Complexity: O(N) where N is the number of nodes in a binary tree.
- Auxiliary Space: O(N)Â
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