Given a binary tree, the task is to print left and right leaf nodes separately.
Examples:
Input:
0
/ \
1 2
/ \
3 4
Output:
Left Leaf Nodes: 3
Right Leaf Nodes: 4 2
Input:
0
\
1
\
2
\
3
Output:
Left Leaf Nodes: None
Right Leaf Nodes: 3
Approach:
- Check if given node is null. If null, then return from the function.
- For each traversal at right and left, send information about the child (left or right child) using the parameter type. Set type = 0 while descending to the left branch and set type = 1 for the right branch.
- Check if it is a leaf node. If the node is a leaf node, then store the leaf node in one of the two vectors of left and right child.
- If node is not a leaf node continue traversal.
- In the case of a single node tree, it will be both a root and a leaf node. This case has to be handled separately.
Below is the implementation of the above approach:
C++
// C++ program for the// above approach#include <bits/stdc++.h>using namespace std;// Structure for// Binary Tree Nodestruct Node { int data; Node* left; Node* right; Node(int x) : data(x) , left(NULL) , right(NULL) { }};// Function for// dfs traversalvoid dfs(Node* root, int type, vector<int>& left_leaf, vector<int>& right_leaf){ // If node is // null, return if (!root) { return; } // If tree consists // of a single node if (!root->left && !root->right) { if (type == -1) { cout << "Tree consists of a single node\n"; } else if (type == 0) { left_leaf.push_back(root->data); } else { right_leaf.push_back(root->data); } return; } // If left child exists, // traverse and set type to 0 if (root->left) { dfs(root->left, 0, left_leaf, right_leaf); } // If right child exists, // traverse and set type to 1 if (root->right) { dfs(root->right, 1, left_leaf, right_leaf); }}// Function to print// the solutionvoid print(vector<int>& left_leaf, vector<int>& right_leaf){ if (left_leaf.size() == 0 && right_leaf.size() == 0) return; // Printing left leaf nodes cout << "Left leaf nodes\n"; for (int x : left_leaf) { cout << x << " "; } cout << '\n'; // Printing right leaf nodes cout << "Right leaf nodes\n"; for (int x : right_leaf) { cout << x << " "; } cout << '\n';}// Driver codeint main(){ Node* root = new Node(0); root->left = new Node(1); root->right = new Node(2); root->left->left = new Node(3); root->left->right = new Node(4); vector<int> left_leaf, right_leaf; dfs(root, -1, left_leaf, right_leaf); print(left_leaf, right_leaf); return 0;} |
Java
// Java program for the// above approachimport java.util.*;class GFG { // Structure for // Binary Tree Node static class Node { int data; Node left; Node right; public Node(int data) { this.data = data; this.left = null; this.right = null; } }; // Function for // dfs traversal static void dfs(Node root, int type, Vector<Integer> left_leaf, Vector<Integer> right_leaf) { // If node is // null, return if (root == null) { return; } // If tree consists // of a single node if (root.left == null && root.right == null) { if (type == -1) { System.out.print("Tree consists of a single node\n"); } else if (type == 0) { left_leaf.add(root.data); } else { right_leaf.add(root.data); } return; } // If left child exists, // traverse and set type to 0 if (root.left != null) { dfs(root.left, 0, left_leaf, right_leaf); } // If right child exists, // traverse and set type to 1 if (root.right != null) { dfs(root.right, 1, left_leaf, right_leaf); } } // Function to print // the solution static void print(Vector<Integer> left_leaf, Vector<Integer> right_leaf) { if (left_leaf.size() == 0 && right_leaf.size() == 0) return; // Printing left leaf nodes System.out.print("Left leaf nodes\n"); for (int x : left_leaf) { System.out.print(x + " "); } System.out.println(); // Printing right leaf nodes System.out.print("Right leaf nodes\n"); for (int x : right_leaf) { System.out.print(x + " "); } System.out.println(); } // Driver code public static void main(String[] args) { Node root = new Node(0); root.left = new Node(1); root.right = new Node(2); root.left.left = new Node(3); root.left.right = new Node(4); Vector<Integer> left_leaf = new Vector<Integer>(), right_leaf = new Vector<Integer>(); dfs(root, -1, left_leaf, right_leaf); print(left_leaf, right_leaf); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program for the# above approach # Structure for# Binary Tree Nodeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # Function for # dfs traversaldef dfs(root, type_t, left_leaf, right_leaf): # If node is # null, return if (not root): return # If tree consists # of a single node if (not root.left and not root.right): if (type_t == -1): print("Tree consists of a single node") elif (type_t == 0): left_leaf.append(root.data) else: right_leaf.append(root.data) return # If left child exists, # traverse and set type_t to 0 if (root.left): dfs(root.left, 0, left_leaf, right_leaf) # If right child exists, # traverse and set type_t to 1 if (root.right): dfs(root.right, 1, left_leaf, right_leaf) # Function to print# the solutiondef prints(left_leaf, right_leaf): if (len(left_leaf) == 0 and len(right_leaf) == 0): return # Printing left leaf nodes print("Left leaf nodes") for x in left_leaf: print(x, end = ' ') print() # Printing right leaf nodes print("Right leaf nodes") for x in right_leaf: print(x, end = ' ') print() # Driver codeif __name__=='__main__': root = Node(0) root.left = Node(1) root.right = Node(2) root.left.left = Node(3) root.left.right = Node(4) left_leaf = [] right_leaf = [] dfs(root, -1, left_leaf, right_leaf) prints(left_leaf, right_leaf) # This code is contributed by pratham76 |
C#
// C# program for the// above approachusing System;using System.Collections.Generic;class GFG { // Structure for // Binary Tree Node public class Node { public int data; public Node left; public Node right; public Node(int data) { this.data = data; this.left = null; this.right = null; } }; // Function for // dfs traversal static void dfs(Node root, int type, List<int> left_leaf, List<int> right_leaf) { // If node is // null, return if (root == null) { return; } // If tree consists // of a single node if (root.left == null && root.right == null) { if (type == -1) { Console.Write("Tree consists of a single node\n"); } else if (type == 0) { left_leaf.Add(root.data); } else { right_leaf.Add(root.data); } return; } // If left child exists, // traverse and set type to 0 if (root.left != null) { dfs(root.left, 0, left_leaf, right_leaf); } // If right child exists, // traverse and set type to 1 if (root.right != null) { dfs(root.right, 1, left_leaf, right_leaf); } } // Function to print // the solution static void print(List<int> left_leaf, List<int> right_leaf) { if (left_leaf.Count == 0 && right_leaf.Count == 0) return; // Printing left leaf nodes Console.Write("Left leaf nodes\n"); foreach (int x in left_leaf) { Console.Write(x + " "); } Console.WriteLine(); // Printing right leaf nodes Console.Write("Right leaf nodes\n"); foreach (int x in right_leaf) { Console.Write(x + " "); } Console.WriteLine(); } // Driver code public static void Main(String[] args) { Node root = new Node(0); root.left = new Node(1); root.right = new Node(2); root.left.left = new Node(3); root.left.right = new Node(4); List<int> left_leaf = new List<int>(), right_leaf = new List<int>(); dfs(root, -1, left_leaf, right_leaf); print(left_leaf, right_leaf); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // JavaScript program for the// above approach// Structure for// Binary Tree Nodeclass Node { constructor(data) { this.data = data; this.left = null; this.right = null; }};// Function for// dfs traversalfunction dfs(root, type, left_leaf, right_leaf) { // If node is // null, return if (root == null) { return; } // If tree consists // of a single node if (root.left == null && root.right == null) { if (type == -1) { document.write("Tree consists of a single node<br>"); } else if (type == 0) { left_leaf.push(root.data); } else { right_leaf.push(root.data); } return; } // If left child exists, // traverse and set type to 0 if (root.left != null) { dfs(root.left, 0, left_leaf, right_leaf); } // If right child exists, // traverse and set type to 1 if (root.right != null) { dfs(root.right, 1, left_leaf, right_leaf); }}// Function to print// the solutionfunction print(left_leaf, right_leaf) { if (left_leaf.Count == 0 && right_leaf.Count == 0) return; // Printing left leaf nodes document.write("Left leaf nodes<br>"); for(var x of left_leaf) { document.write(x + " "); } document.write("<br>"); // Printing right leaf nodes document.write("Right leaf nodes<br>"); for(var x of right_leaf) { document.write(x + " "); } document.write("<br>");}// Driver codevar root = new Node(0);root.left = new Node(1);root.right = new Node(2);root.left.left = new Node(3);root.left.right = new Node(4);var left_leaf = [];var right_leaf = [];dfs(root, -1, left_leaf, right_leaf);print(left_leaf, right_leaf);</script> |
Output:
Left leaf nodes 3 Right leaf nodes 4 2
Time complexity: O(h) where h is height of given Binary Tree
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