Find sum of all left leaves in a given Binary Tree
Given a Binary Tree, find the sum of all left leaves in it. For example, sum of all left leaves in below Binary Tree is 5+1=6.
The idea is to traverse the tree, starting from root. For every node, check if its left subtree is a leaf. If it is, then add it to the result.
Following is the implementation of the above idea.
C++
// A C++ program to find sum of all left leaves#include <bits/stdc++.h>using namespace std;/* A binary tree Node has key, pointer to left and right children */struct Node{ int key; struct Node* left, *right;};/* Helper function that allocates a new node with the given data and NULL left and right pointer. */Node *newNode(char k){ Node *node = new Node; node->key = k; node->right = node->left = NULL; return node;}// A utility function to check if a given node is leaf or notbool isLeaf(Node *node){ if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return true; return false;}// This function returns sum of all left leaves in a given// binary treeint leftLeavesSum(Node *root){ // Initialize result int res = 0; // Update result if root is not NULL if (root != NULL) { // If left of root is NULL, then add key of // left child if (isLeaf(root->left)) res += root->left->key; else // Else recur for left child of root res += leftLeavesSum(root->left); // Recur for right child of root and update res res += leftLeavesSum(root->right); } // return result return res;}/* Driver program to test above functions*/int main(){ // Let us a construct the Binary Tree struct Node *root = newNode(20); root->left = newNode(9); root->right = newNode(49); root->right->left = newNode(23); root->right->right = newNode(52); root->right->right->left = newNode(50); root->left->left = newNode(5); root->left->right = newNode(12); root->left->right->right = newNode(12); cout << "Sum of left leaves is " << leftLeavesSum(root); return 0;} |
Java
// Java program to find sum of all left leavesclass Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }} class BinaryTree { Node root; // A utility function to check if a given node is leaf or not boolean isLeaf(Node node) { if (node == null) return false; if (node.left == null && node.right == null) return true; return false; } // This function returns sum of all left leaves in a given // binary tree int leftLeavesSum(Node node) { // Initialize result int res = 0; // Update result if root is not NULL if (node != null) { // If left of root is NULL, then add key of // left child if (isLeaf(node.left)) res += node.left.data; else // Else recur for left child of root res += leftLeavesSum(node.left); // Recur for right child of root and update res res += leftLeavesSum(node.right); } // return result return res; } // Driver program public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(9); tree.root.right = new Node(49); tree.root.left.right = new Node(12); tree.root.left.left = new Node(5); tree.root.right.left = new Node(23); tree.root.right.right = new Node(52); tree.root.left.right.right = new Node(12); tree.root.right.right.left = new Node(50); System.out.println("The sum of leaves is " + tree.leftLeavesSum(tree.root)); }} // This code is contributed by Mayank Jaiswal |
Python
# Python program to find sum of all left leaves# A Binary tree nodeclass Node: # Constructor to create a new Node def __init__(self, key): self.key = key self.left = None self.right = None# A utility function to check if a given node is leaf or notdef isLeaf(node): if node is None: return False if node.left is None and node.right is None: return True return False# This function return sum of all left leaves in a# given binary treedef leftLeavesSum(root): # Initialize result res = 0 # Update result if root is not None if root is not None: # If left of root is None, then add key of # left child if isLeaf(root.left): res += root.left.key else: # Else recur for left child of root res += leftLeavesSum(root.left) # Recur for right child of root and update res res += leftLeavesSum(root.right) return res# Driver program to test above function # Let us constrcut the Binary Tree shown in the above functionroot = Node(20)root.left = Node(9)root.right = Node(49)root.right.left = Node(23) root.right.right = Node(52)root.right.right.left = Node(50)root.left.left = Node(5)root.left.right = Node(12)root.left.right.right = Node(12)print "Sum of left leaves is", leftLeavesSum(root)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// C# program to find sum of all left leaves public class Node{ public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class BinaryTree{ public Node root; // A utility function to check if a given node is leaf or not public virtual bool isLeaf(Node node) { if (node == null) { return false; } if (node.left == null && node.right == null) { return true; } return false; } // This function returns sum of all left leaves in a given // binary tree public virtual int leftLeavesSum(Node node) { // Initialize result int res = 0; // Update result if root is not NULL if (node != null) { // If left of root is NULL, then add key of // left child if (isLeaf(node.left)) { res += node.left.data; } else // Else recur for left child of root { res += leftLeavesSum(node.left); } // Recur for right child of root and update res res += leftLeavesSum(node.right); } // return result return res; } // Driver program public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(9); tree.root.right = new Node(49); tree.root.left.right = new Node(12); tree.root.left.left = new Node(5); tree.root.right.left = new Node(23); tree.root.right.right = new Node(52); tree.root.left.right.right = new Node(12); tree.root.right.right.left = new Node(50); Console.WriteLine("The sum of leaves is " + tree.leftLeavesSum(tree.root)); }} // This code is contributed by Shrikant13 |
Output
Sum of left leaves is 78
Time Complexity: O(N), where n is number of nodes in Binary Tree.
Following is Another Method to solve the above problem. This solution passes in a sum variable as an accumulator. When a left leaf is encountered, the leaf’s data is added to sum. Time complexity of this method is also O(n). Thanks to Xin Tong (geeksforgeeks userid trent.tong) for suggesting this method.
C++
// A C++ program to find sum of all left leaves#include <bits/stdc++.h>using namespace std;/* A binary tree Node has key, pointer to left and right children */struct Node{ int key; struct Node* left, *right;};/* Helper function that allocates a new node with the given data and NULL left and right pointer. */Node *newNode(char k){ Node *node = new Node; node->key = k; node->right = node->left = NULL; return node;}/* Pass in a sum variable as an accumulator */void leftLeavesSumRec(Node *root, bool isleft, int *sum){ if (!root) return; // Check whether this node is a leaf node and is left. if (!root->left && !root->right && isleft) *sum += root->key; // Pass 1 for left and 0 for right leftLeavesSumRec(root->left, 1, sum); leftLeavesSumRec(root->right, 0, sum);}// A wrapper over above recursive functionint leftLeavesSum(Node *root){ int sum = 0; //Initialize result // use the above recursive function to evaluate sum leftLeavesSumRec(root, 0, &sum); return sum;}/* Driver program to test above functions*/int main(){ // Let us construct the Binary Tree shown in the // above figure int sum = 0; struct Node *root = newNode(20); root->left = newNode(9); root->right = newNode(49); root->right->left = newNode(23); root->right->right = newNode(52); root->right->right->left = newNode(50); root->left->left = newNode(5); root->left->right = newNode(12); root->left->right->right = newNode(12); cout << "Sum of left leaves is " << leftLeavesSum(root) << endl; return 0;} |
Java
// Java program to find sum of all left leavesclass Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }} // Passing sum as accumulator and implementing pass by reference // of sum variable class Sum { int sum = 0;} class BinaryTree { Node root; /* Pass in a sum variable as an accumulator */ void leftLeavesSumRec(Node node, boolean isleft, Sum summ) { if (node == null) return; // Check whether this node is a leaf node and is left. if (node.left == null && node.right == null && isleft) summ.sum = summ.sum + node.data; // Pass true for left and false for right leftLeavesSumRec(node.left, true, summ); leftLeavesSumRec(node.right, false, summ); } // A wrapper over above recursive function int leftLeavesSum(Node node) { Sum suum = new Sum(); // use the above recursive function to evaluate sum leftLeavesSumRec(node, false, suum); return suum.sum; } // Driver program public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(9); tree.root.right = new Node(49); tree.root.left.right = new Node(12); tree.root.left.left = new Node(5); tree.root.right.left = new Node(23); tree.root.right.right = new Node(52); tree.root.left.right.right = new Node(12); tree.root.right.right.left = new Node(50); System.out.println("The sum of leaves is " + tree.leftLeavesSum(tree.root)); }} // This code is contributed by Mayank Jaiswal |
Python
# Python program to find sum of all left leaves# A binary tree nodeclass Node: # A constructor to create a new Node def __init__(self, key): self.key = key self.left = None self.right = Nonedef leftLeavesSumRec(root, isLeft, summ): if root is None: return # Check whether this node is a leaf node and is left if root.left is None and root.right is None and isLeft == True: summ[0] += root.key # Pass 1 for left and 0 for right leftLeavesSumRec(root.left, 1, summ) leftLeavesSumRec(root.right, 0, summ) # A wrapper over above recursive functiondef leftLeavesSum(root): summ = [0] # initialize result # Use the above recursive fucntion to evaluate sum leftLeavesSumRec(root, 0, summ) return summ[0]# Driver program to test above function# Let us construct the Binary Tree shown in the# above figureroot = Node(20);root.left= Node(9);root.right = Node(49);root.right.left = Node(23);root.right.right= Node(52);root.right.right.left = Node(50);root.left.left = Node(5);root.left.right = Node(12);root.left.right.right = Node(12);print "Sum of left leaves is", leftLeavesSum(root) # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// C# program to find sum of all left leaves public class Node{ public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}// Passing sum as accumulator and implementing pass by reference // of sum variable public class Sum{ public int sum = 0;}public class BinaryTree{ public Node root; /* Pass in a sum variable as an accumulator */ public virtual void leftLeavesSumRec(Node node, bool isleft, Sum summ) { if (node == null) { return; } // Check whether this node is a leaf node and is left. if (node.left == null && node.right == null && isleft) { summ.sum = summ.sum + node.data; } // Pass true for left and false for right leftLeavesSumRec(node.left, true, summ); leftLeavesSumRec(node.right, false, summ); } // A wrapper over above recursive function public virtual int leftLeavesSum(Node node) { Sum suum = new Sum(); // use the above recursive function to evaluate sum leftLeavesSumRec(node, false, suum); return suum.sum; } // Driver program public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(20); tree.root.left = new Node(9); tree.root.right = new Node(49); tree.root.left.right = new Node(12); tree.root.left.left = new Node(5); tree.root.right.left = new Node(23); tree.root.right.right = new Node(52); tree.root.left.right.right = new Node(12); tree.root.right.right.left = new Node(50); Console.WriteLine("The sum of leaves is " + tree.leftLeavesSum(tree.root)); }} // This code is contributed by Shrikant13 |
Output
Sum of left leaves is 78
Iterative Approach :
This is the Iterative Way to find the sum of the left leaves.
Idea is to perform Depth-First Traversal on the tree (either Inorder, Preorder or Postorder) using a stack and checking if the Left Child is a Leaf node. If it is, then add the nodes value to the sum variable
C++
// C++ program to find sum of all left leaves#include<bits/stdc++.h>using namespace std;// A binary tree nodeclass Node{ public: int key; Node* left, *right; // A constructor to create a new Node Node(int key_) { key = key_; left = NULL; right = NULL; }};// Return the sum of left leaf nodesint sumOfLeftLeaves(Node* root){ if(root == NULL) return 0; // Using a stack_ for Depth-First // Traversal of the tree stack<Node*> stack_; stack_.push(root); // sum holds the sum of all the left leaves int sum = 0; while(stack_.size() > 0) { Node* currentNode = stack_.top(); stack_.pop(); if (currentNode->left != NULL) { stack_.push(currentNode->left); // Check if currentNode's left // child is a leaf node if(currentNode->left->left == NULL && currentNode->left->right == NULL) { // if currentNode is a leaf, // add its data to the sum sum = sum + currentNode->left->key ; } } if (currentNode->right != NULL) stack_.push(currentNode->right); } return sum;}// Driver Codeint main(){ Node *root = new Node(20); root->left= new Node(9); root->right = new Node(49); root->right->left = new Node(23); root->right->right= new Node(52); root->right->right->left = new Node(50); root->left->left = new Node(5); root->left->right = new Node(12); root->left->right->right = new Node(12); cout << "Sum of left leaves is " << sumOfLeftLeaves(root) << endl; return 0;}// This code is contributed by Arnab Kundu |
Java
// Java program to find sum of all left leavesimport java.util.*;class GFG{// A binary tree nodestatic class Node{ int key; Node left, right; // A constructor to create a new Node Node(int key_) { this.key = key_; this.left = null; this.right = null; }};// Return the sum of left leaf nodesstatic int sumOfLeftLeaves(Node root){ if(root == null) return 0; // Using a stack_ for Depth-First // Traversal of the tree Stack<Node> stack_ = new Stack<>(); stack_.push(root); // sum holds the sum of all the left leaves int sum = 0; while(stack_.size() > 0) { Node currentNode = stack_.peek(); stack_.pop(); if (currentNode.left != null) { stack_.add(currentNode.left); // Check if currentNode's left // child is a leaf node if(currentNode.left.left == null && currentNode.left.right == null) { // if currentNode is a leaf, // add its data to the sum sum = sum + currentNode.left.key ; } } if (currentNode.right != null) stack_.add(currentNode.right); } return sum;}// Driver Codepublic static void main(String[] args){ Node root = new Node(20); root.left= new Node(9); root.right = new Node(49); root.right.left = new Node(23); root.right.right= new Node(52); root.right.right.left = new Node(50); root.left.left = new Node(5); root.left.right = new Node(12); root.left.right.right = new Node(12); System.out.print("Sum of left leaves is " + sumOfLeftLeaves(root) +"\n");}}// This code is contributed by aashish1995 |
Python3
# Python3 program to find sum of all left leaves# A binary tree nodeclass Node: # A constructor to create a new Node def __init__(self, key): self.key = key self.left = None self.right = None# Return the sum of left leaf nodesdef sumOfLeftLeaves(root): if(root is None): return # Using a stack for Depth-First Traversal of the tree stack = [] stack.append(root) # sum holds the sum of all the left leaves sum = 0 while len(stack) > 0: currentNode = stack.pop() if currentNode.left is not None: stack.append(currentNode.left) # Check if currentNode's left child is a leaf node if currentNode.left.left is None and currentNode.left.right is None: # if currentNode is a leaf, add its data to the sum sum = sum + currentNode.left.data if currentNode.right is not None: stack.append(currentNode.right) return sum# Driver Coderoot = Tree(20);root.left= Tree(9);root.right = Tree(49);root.right.left = Tree(23);root.right.right= Tree(52);root.right.right.left = Tree(50);root.left.left = Tree(5);root.left.right = Tree(12);root.left.right.right = Tree(12);print('Sum of left leaves is {}'.format(sumOfLeftLeaves(root))) |
C#
// C# program to find sum of all left leavesusing System;using System.Collections.Generic;class GFG{ // A binary tree node public class Node { public int key; public Node left, right; // A constructor to create a new Node public Node(int key_) { this.key = key_; this.left = null; this.right = null; } }; // Return the sum of left leaf nodes static int sumOfLeftLeaves(Node root) { if(root == null) return 0; // Using a stack_ for Depth-First // Traversal of the tree Stack<Node> stack_ = new Stack<Node>(); stack_.Push(root); // sum holds the sum of all the left leaves int sum = 0; while(stack_.Count > 0) { Node currentNode = stack_.Peek(); stack_.Pop(); if (currentNode.left != null) { stack_.Push(currentNode.left); // Check if currentNode's left // child is a leaf node if(currentNode.left.left == null && currentNode.left.right == null) { // if currentNode is a leaf, // add its data to the sum sum = sum + currentNode.left.key ; } } if (currentNode.right != null) stack_.Push(currentNode.right); } return sum; } // Driver Code public static void Main(String[] args) { Node root = new Node(20); root.left= new Node(9); root.right = new Node(49); root.right.left = new Node(23); root.right.right= new Node(52); root.right.right.left = new Node(50); root.left.left = new Node(5); root.left.right = new Node(12); root.left.right.right = new Node(12); Console.Write("Sum of left leaves is " + sumOfLeftLeaves(root) +"\n"); }}// This code is contributed by Rajput-Ji |
Output
Sum of left leaves is 78
Thanks to Shubham Tambere for suggesting this approach.
BFS Approach: We can do BFS traversal and keep a separate variable for denoting if it is a left child or right child of a node. As soon as we encounter a leaf, we check if it is a left child of its parent or right child of its parent. If it is a left child, we add its value in the sum.
Below is the implementation of the above approach:
C++
// C++ program to find sum of all left leaves#include <bits/stdc++.h>using namespace std;// A binary tree nodeclass Node {public: int key; Node *left, *right; // constructor to create a new Node Node(int key_) { key = key_; left = NULL; right = NULL; }};// Return the sum of left leaf nodesint sumOfLeftLeaves(Node* root){ if (root == NULL) return 0; // A queue of pairs to do bfs traversal // and keep track if the node is a left // or right child if boolean value // is true then it is a left child. queue<pair<Node*, bool> > q; q.push({ root, 0 }); int sum = 0; // do bfs traversal while (!q.empty()) { Node* temp = q.front().first; bool is_left_child = q.front().second; q.pop(); // if temp is a leaf node and // left child of its parent if (!temp->left && !temp->right && is_left_child) sum = sum + temp->key; // if it is not leaf then // push its children nodes // into queue if (temp->left) { // boolean value is true // here because it is left // child of its parent q.push({ temp->left, 1 }); } if (temp->right) { // boolean value is false // here because it is // right child of its parent q.push({ temp->right, 0 }); } } return sum;}// Driver Codeint main(){ Node* root = new Node(20); root->left = new Node(9); root->right = new Node(49); root->right->left = new Node(23); root->right->right = new Node(52); root->right->right->left = new Node(50); root->left->left = new Node(5); root->left->right = new Node(12); root->left->right->right = new Node(12); cout << "Sum of left leaves is " << sumOfLeftLeaves(root) << endl; return 0;} |
Java
// Java program to find sum of all left leavesimport java.util.*;class GFG{// A binary tree nodestatic class Node { int key; Node left, right; // constructor to create a new Node Node(int key_) { key = key_; left = null; right = null; }};static class pair{ Node first; boolean second; public pair(Node first, boolean second) { this.first = first; this.second = second; } } // Return the sum of left leaf nodesstatic int sumOfLeftLeaves(Node root){ if (root == null) return 0; // A queue of pairs to do bfs traversal // and keep track if the node is a left // or right child if boolean value // is true then it is a left child. Queue<pair > q = new LinkedList<>(); q.add(new pair( root, false )); int sum = 0; // do bfs traversal while (!q.isEmpty()) { Node temp = q.peek().first; boolean is_left_child = q.peek().second; q.remove(); // if temp is a leaf node and // left child of its parent if (is_left_child) sum = sum + temp.key; if(temp.left != null && temp.right != null && is_left_child) sum = sum-temp.key; // if it is not leaf then // push its children nodes // into queue if (temp.left != null) { // boolean value is true // here because it is left // child of its parent q.add(new pair( temp.left, true)); } if (temp.right != null) { // boolean value is false // here because it is // right child of its parent q.add(new pair( temp.right, false)); } } return sum;}// Driver Codepublic static void main(String[] args){ Node root = new Node(20); root.left = new Node(9); root.right = new Node(49); root.right.left = new Node(23); root.right.right = new Node(52); root.right.right.left = new Node(50); root.left.left = new Node(5); root.left.right = new Node(12); root.left.right.right = new Node(12); System.out.print("Sum of left leaves is " + sumOfLeftLeaves(root) +"\n");}}// This code is contributed by gauravrajput1. |
Python3
# Python3 program to find sum of# all left leavesfrom collections import deque# A binary tree nodeclass Node: def __init__(self, x): self.key = x self.left = None self.right = None# Return the sum of left leaf nodesdef sumOfLeftLeaves(root): if (root == None): return 0 # A queue of pairs to do bfs traversal # and keep track if the node is a left # or right child if boolean value # is true then it is a left child. q = deque() q.append([root, 0]) sum = 0 # Do bfs traversal while (len(q) > 0): temp = q[0][0] is_left_child = q[0][1] q.popleft() # If temp is a leaf node and # left child of its parent if (not temp.left and not temp.right and is_left_child): sum = sum + temp.key # If it is not leaf then # push its children nodes # into queue if (temp.left): # Boolean value is true # here because it is left # child of its parent q.append([temp.left, 1]) if (temp.right): # Boolean value is false # here because it is # right child of its parent q.append([temp.right, 0]) return sum# Driver Codeif __name__ == '__main__': root = Node(20) root.left = Node(9) root.right = Node(49) root.right.left = Node(23) root.right.right = Node(52) root.right.right.left = Node(50) root.left.left = Node(5) root.left.right = Node(12) root.left.right.right = Node(12) print("Sum of left leaves is", sumOfLeftLeaves(root))# This code is contributed by mohit kumar 29 |
Output
Sum of left leaves is 78
Time Complexity: O(N)
Auxiliary Space: O(N)
This article is contributed by Manish. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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