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 same as the given level, then check whether the current node is leaf node or not. If it is a leaf node then print it.
Below is the implementation of the above approach:
C++
// C++ program to print all the // leaf nodes at a given level // in a Binary tree #include <bits/stdc++.h> using namespace std; // Binary tree node struct node { struct node* left; struct node* right; int data; }; // Function to create a new // Binary node struct node* newNode(int data) { struct node* temp = new node; temp->data = data; temp->left = NULL; temp->right = NULL; return temp; } // Function to print Leaf Nodes at // a given level void PrintLeafNodes(struct node* root, int level) { if (root == NULL) return; 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); } } // Driver code int main() { struct 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); int level = 4; PrintLeafNodes(root, level); return 0; } |
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Java
// Java program to print all the // leaf nodes at a given level // in a Binary tree class GFG { // Binary tree node static class node { node left; node right; int data; }; // Function to create a new // Binary node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Function to print Leaf Nodes at // a given level static void PrintLeafNodes(node root, int level) { if (root == null) { return; } if (level == 1) { if (root.left == null && root.right == null) { System.out.print(root.data + " "); } } else if (level > 1) { PrintLeafNodes(root.left, level - 1); PrintLeafNodes(root.right, level - 1); } } // 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); int level = 4; PrintLeafNodes(root, level); } } // This code contributed by Rajput-Ji |
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C#
// C# program to print all the // leaf nodes at a given level // in a Binary tree using System; class GFG { // Binary tree node public class node { public node left; public node right; public int data; }; // Function to create a new // Binary node static node newNode(int data) { node temp = new node(); temp.data = data; temp.left = null; temp.right = null; return temp; } // Function to print Leaf Nodes at // a given level static void PrintLeafNodes(node root, int level) { if (root == null) { return; } if (level == 1) { if (root.left == null && root.right == null) { Console.Write(root.data + " "); } } else if (level > 1) { PrintLeafNodes(root.left, level - 1); PrintLeafNodes(root.right, level - 1); } } // 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); int level = 4; PrintLeafNodes(root, level); } } // This code has been contributed by 29AjayKumar |
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Output:
8 7
Time Complexity:O(N) where N is number of nodes in a binary tree
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