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Find sum of all left leaves in a given Binary Tree

Given a Binary Tree, find sum of all left leaves in it. For example, sum of all left leaves in below Binary Tree is 5+1=6.

tree



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 above idea.

C++

// A C++ program to find sum of all left leaves
#include <iostream>
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 not
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
int 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 leaves
class 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 node
class 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 not
def 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 tree
def 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 function
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", 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 of the above solution is 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 <iostream>
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 function
int 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 leaves
class 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 node
class Node:
  
    # A constructor to create a new Node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
  
def 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 function
def 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 figure
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", 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

# Python program to find sum of all left leaves
  
# A binary tree node
class 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 nodes
def 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 Code
root = 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)))

Output: 

Sum of left leaves is 78

Thanks to Shubham Tambere for suggesting this approach.



This article is attributed to GeeksforGeeks.org

tags:

Tree Traversal Traversal Tree
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