Skip to content
Related Articles

Related Articles

Sum of nodes at maximum depth of a Binary Tree
  • Difficulty Level : Easy
  • Last Updated : 21 Apr, 2021

Given a root node to a tree, find the sum of all the leaf nodes which are at maximum depth from root node.
Example: 
 

      1
    /   \
   2     3
  / \   / \
 4   5 6   7

Input : root(of above tree)
Output : 22

Explanation:
Nodes at maximum depth are: 4, 5, 6, 7. 
So, sum of these nodes = 22

 

While traversing the nodes compare the level of the node with max_level (Maximum level till the current node). If the current level exceeds the maximum level, update the max_level as the current level. If the max level and current level are the same, add the root data to the current sum. if the level is less than max_level, do nothing. 
 

C++




// Code to find the sum of the nodes
// which are present at the maximum depth.
#include <bits/stdc++.h>
 
using namespace std;
 
int sum = 0, max_level = INT_MIN;
 
struct Node
{
    int d;
    Node *l;
    Node *r;
};
 
// Function to return a new node
Node* createNode(int d)
{
    Node *node;
    node = new Node;
    node->d = d;
    node->l = NULL;
    node->r = NULL;
    return node;
}
 
// Function to find the sum of the node
// which are present at the maximum depth.
// While traversing the nodes compare the level
// of the node with max_level
// (Maximum level till the current node).
// If the current level exceeds the maximum level,
// update the max_level as current level.
// If the max level and current level are same,
// add the root data to current sum.
void sumOfNodesAtMaxDepth(Node *ro,int level)
{
    if(ro == NULL)
    return;
    if(level > max_level)
    {
        sum = ro -> d;
        max_level = level;
    }
    else if(level == max_level)
    {
        sum = sum + ro -> d;
    }
    sumOfNodesAtMaxDepth(ro -> l, level + 1);
    sumOfNodesAtMaxDepth(ro -> r, level + 1);
}
 
// Driver Code
int main()
{
    Node *root;
    root = createNode(1);
    root->l = createNode(2);
    root->r = createNode(3);
    root->l->l = createNode(4);
    root->l->r = createNode(5);
    root->r->l = createNode(6);
    root->r->r = createNode(7);
    sumOfNodesAtMaxDepth(root, 0);
    cout << sum;
    return 0;
}

Java




// Java code to find the sum of the nodes
// which are present at the maximum depth.
class GFG
{
 
static int sum = 0, max_level = Integer.MIN_VALUE;
 
static class Node
{
    int d;
    Node l;
    Node r;
};
 
// Function to return a new node
static Node createNode(int d)
{
    Node node;
    node = new Node();
    node.d = d;
    node.l = null;
    node.r = null;
    return node;
}
 
// Function to find the sum of the node
// which are present at the maximum depth.
// While traversing the nodes compare the level
// of the node with max_level
// (Maximum level till the current node).
// If the current level exceeds the maximum level,
// update the max_level as current level.
// If the max level and current level are same,
// add the root data to current sum.
static void sumOfNodesAtMaxDepth(Node ro,int level)
{
    if(ro == null)
    return;
    if(level > max_level)
    {
        sum = ro . d;
        max_level = level;
    }
    else if(level == max_level)
    {
        sum = sum + ro . d;
    }
    sumOfNodesAtMaxDepth(ro . l, level + 1);
    sumOfNodesAtMaxDepth(ro . r, level + 1);
}
 
// Driver Code
public static void main(String[] args)
{
    Node root;
    root = createNode(1);
    root.l = createNode(2);
    root.r = createNode(3);
    root.l.l = createNode(4);
    root.l.r = createNode(5);
    root.r.l = createNode(6);
    root.r.r = createNode(7);
    sumOfNodesAtMaxDepth(root, 0);
    System.out.println(sum);
}
}
 
/* This code is contributed by PrinciRaj1992 */

Python3




# Python3 code to find the sum of the nodes
# which are present at the maximum depth.
sum = [0]
max_level = [-(2**32)]
 
# Binary tree node
class createNode:
     
    def __init__(self, data):
        self.d = data
        self.l = None
        self.r = None
 
# Function to find the sum of the node
# which are present at the maximum depth.
# While traversing the nodes compare the level
# of the node with max_level
# (Maximum level till the current node).
# If the current level exceeds the maximum level,
# update the max_level as current level.
# If the max level and current level are same,
# add the root data to current sum.
def sumOfNodesAtMaxDepth(ro, level):
     
    if(ro == None):
        return
     
    if(level > max_level[0]):
        sum[0] = ro . d
        max_level[0] = level
     
    elif(level == max_level[0]):
        sum[0] = sum[0] + ro . d
     
    sumOfNodesAtMaxDepth(ro . l, level + 1)
    sumOfNodesAtMaxDepth(ro . r, level + 1)
     
# Driver Code
root = createNode(1)
root.l = createNode(2)
root.r = createNode(3)
root.l.l = createNode(4)
root.l.r = createNode(5)
root.r.l = createNode(6)
root.r.r = createNode(7)
sumOfNodesAtMaxDepth(root, 0)
print(sum[0])
 
# This code is contributed by SHUBHAMSINGH10

C#




// C# code to find the sum of the nodes
// which are present at the maximum depth.
using System;
 
class GFG
{
 
static int sum = 0, max_level = int.MinValue;
 
public class Node
{
    public int d;
    public Node l;
    public Node r;
};
 
// Function to return a new node
static Node createNode(int d)
{
    Node node;
    node = new Node();
    node.d = d;
    node.l = null;
    node.r = null;
    return node;
}
 
// Function to find the sum of the node
// which are present at the maximum depth.
// While traversing the nodes compare the level
// of the node with max_level
// (Maximum level till the current node).
// If the current level exceeds the maximum level,
// update the max_level as current level.
// If the max level and current level are same,
// add the root data to current sum.
static void sumOfNodesAtMaxDepth(Node ro,int level)
{
    if(ro == null)
    return;
    if(level > max_level)
    {
        sum = ro . d;
        max_level = level;
    }
    else if(level == max_level)
    {
        sum = sum + ro . d;
    }
    sumOfNodesAtMaxDepth(ro . l, level + 1);
    sumOfNodesAtMaxDepth(ro . r, level + 1);
}
 
// Driver Code
public static void Main(String[] args)
{
    Node root;
    root = createNode(1);
    root.l = createNode(2);
    root.r = createNode(3);
    root.l.l = createNode(4);
    root.l.r = createNode(5);
    root.r.l = createNode(6);
    root.r.r = createNode(7);
    sumOfNodesAtMaxDepth(root, 0);
    Console.WriteLine(sum);
}
}
 
// This code is contributed by Princi Singh

Output : 
 

22

This article is contributed by Ashwin Loganathan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Approach: Calculate the max depth of the given tree. Now, start traversing the tree similarly as traversed during maximum depth calculation. But, this time with one more argument (i.e. maxdepth), and traverse recursively with decreasing depth by 1 for each left or right call. Wherever max == 1, means the node at max depth is reached. So add its data value to sum. Finally, return sum.
Below is the implementation for the above approach: 
 



C++




// C++ code for sum of nodes
// at maximum depth
#include <bits/stdc++.h>
using namespace std;
 
struct Node
{
    int data;
    Node* left, *right;
 
    // Constructor
    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
    // function to find the sum of nodes at
    // maximum depth arguments are node and
    // max, where max is to match the depth
    // of node at every call to node, if
    // max will be equal to 1, means
    // we are at deepest node.
    int sumMaxLevelRec(Node* node, int max)
    {
        // base case
        if (node == NULL)
            return 0;    
 
        // max == 1 to track the node
        // at deepest level
        if (max == 1)
            return node->data;
 
        // recursive call to left and right nodes
        return sumMaxLevelRec(node->left, max - 1) +
            sumMaxLevelRec(node->right, max - 1);
    }
     
    // maxDepth function to find the
    // max depth of the tree
    int maxDepth(Node* node)
    {
        // base case
        if (node == NULL)
            return 0;    
 
        // either leftDepth of rightDepth is
        // greater add 1 to include height
        // of node at which call is
        return 1 + max(maxDepth(node->left),
                        maxDepth(node->right));    
    }
 
    int sumMaxLevel(Node* root)
    {
 
        // call to function to calculate
        // max depth
        int MaxDepth = maxDepth(root);
         
        return sumMaxLevelRec(root, MaxDepth);
    }
 
    // Driver code
    int main()
    {
 
        /*     1
            / \
            2 3
            / \ / \
            4 5 6 7     */
 
        // Constructing tree
        Node* root = new Node(1);
        root->left = new Node(2);
        root->right = new Node(3);
        root->left->left = new Node(4);
        root->left->right = new Node(5);
        root->right->left = new Node(6);
        root->right->right = new Node(7);
 
        // call to calculate required sum
        cout<<(sumMaxLevel(root))<<endl;
    }
 
// This code is contributed by Arnab Kundu

Java




// Java code for sum of nodes
// at maximum depth
import java.util.*;
 
class Node {
    int data;
    Node left, right;
 
    // Constructor
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
class GfG {
 
    // function to find the sum of nodes at
    // maximum depth arguments are node and
    // max, where max is to match the depth
    // of node at every call to node, if
    // max will be equal to 1, means
    // we are at deepest node.
    public static int sumMaxLevelRec(Node node,
                     int max)
    {
        // base case
        if (node == null)
            return 0;    
 
        // max == 1 to track the node
        // at deepest level
        if (max == 1)
            return node.data;   
 
        // recursive call to left and right nodes
        return sumMaxLevelRec(node.left, max - 1) +
               sumMaxLevelRec(node.right, max - 1);
    }
 
    public static int sumMaxLevel(Node root) {
 
        // call to function to calculate
        // max depth
        int MaxDepth = maxDepth(root);
         
        return sumMaxLevelRec(root, MaxDepth);
    }
 
    // maxDepth function to find the
    // max depth of the tree
    public static int maxDepth(Node node)
    {
        // base case
        if (node == null)
            return 0;    
 
        // either leftDepth of rightDepth is
        // greater add 1 to include height
        // of node at which call is
        return 1 + Math.max(maxDepth(node.left),
                           maxDepth(node.right));    
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        /*      1
              / \
              2   3
            / \ / \
            4 5 6 7     */
 
        // Constructing tree
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
 
 
        // call to calculate required sum
        System.out.println(sumMaxLevel(root));
    }
}

Python3




# Python3 code for sum of nodes at maximum depth
class Node:
 
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to find the sum of nodes at maximum depth
# arguments are node and max, where Max is to match
# the depth of node at every call to node, if Max
# will be equal to 1, means we are at deepest node.
def sumMaxLevelRec(node, Max):
 
    # base case
    if node == None:
        return 0   
 
    # Max == 1 to track the node at deepest level
    if Max == 1:
        return node.data    
 
    # recursive call to left and right nodes
    return (sumMaxLevelRec(node.left, Max - 1) +
            sumMaxLevelRec(node.right, Max - 1))
 
def sumMaxLevel(root):
 
    # call to function to calculate max depth
    MaxDepth = maxDepth(root)
    return sumMaxLevelRec(root, MaxDepth)
 
# maxDepth function to find
# the max depth of the tree
def maxDepth(node):
 
    # base case
    if node == None:
        return 0   
 
    # Either leftDepth of rightDepth is
    # greater add 1 to include height
    # of node at which call is
    return 1 + max(maxDepth(node.left),
                    maxDepth(node.right))
 
# Driver code
if __name__ == "__main__":
 
    # Constructing tree
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.left = Node(6)
    root.right.right = Node(7)
 
    # call to calculate required sum
    print(sumMaxLevel(root))
 
# This code is contributed by Rituraj Jain

C#




using System;
 
// C# code for sum of nodes
// at maximum depth
 
public class Node
{
    public int data;
    public Node left, right;
 
    // Constructor
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
public class GfG
{
 
    // function to find the sum of nodes at
    // maximum depth arguments are node and
    // max, where max is to match the depth
    // of node at every call to node, if
    // max will be equal to 1, means
    // we are at deepest node.
    public static int sumMaxLevelRec(Node node, int max)
    {
        // base case
        if (node == null)
        {
            return 0;
        }
 
        // max == 1 to track the node
        // at deepest level
        if (max == 1)
        {
            return node.data;
        }
 
        // recursive call to left and right nodes
        return sumMaxLevelRec(node.left, max - 1)
                + sumMaxLevelRec(node.right, max - 1);
    }
 
    public static int sumMaxLevel(Node root)
    {
 
        // call to function to calculate
        // max depth
        int MaxDepth = maxDepth(root);
 
        return sumMaxLevelRec(root, MaxDepth);
    }
 
    // maxDepth function to find the
    // max depth of the tree
    public static int maxDepth(Node node)
    {
        // base case
        if (node == null)
        {
            return 0;
        }
 
        // either leftDepth of rightDepth is
        // greater add 1 to include height
        // of node at which call is
        return 1 + Math.Max(maxDepth(node.left),
                            maxDepth(node.right));
    }
 
    // Driver code
    public static void Main(string[] args)
    {
 
        /*     1
            / \
            2 3
            / \ / \
            4 5 6 7     */
 
        // Constructing tree
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
 
 
        // call to calculate required sum
        Console.WriteLine(sumMaxLevel(root));
    }
}
 
// This code is contributed by Shrikant13

Output : 
 

22

Time Complexity: O(N), where N is the number of nodes in the tree.

Method 3:- Using a Queue

Approach: In this approach, the idea is to traverse the tree in level order and for each level calculate the sum of all the nodes in that level. For each level, we push all the nodes of the tree into the queue and calculate the sum of the nodes. So, when we reach the end leaves in the tree, the sum is the total sum of all the leaves in the binary tree.

Below is the implementation of the above approach: 

Java




// Java code to print the sum of leaves present at maximum
// depth of a binary tree
import java.util.*;
 
class GFG {
 
    static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
    };
 
    // Function to return a new node
    static TreeNode createNode(int d)
    {
        TreeNode node;
        node = new TreeNode();
        node.val = d;
        node.left = null;
        node.right = null;
        return node;
    }
    // Iterative function to fnd the sum of the deepest
    // nodes.
    public static int deepestLeavesSum(TreeNode root)
    {
          // if the root is null then return 0
        if (root== null) {
            return 0;
        }
          // Initialize an empty queue.
        Queue<TreeNode> qu= new LinkedList<>();
        // push the root of the tree into the queue
        qu.offer(root);
        // initialize sum of current level to 0
        int sumOfCurrLevel= 0;
          // loop until the queue is not empty
        while (!qu.isEmpty()) {
            int size = qu.size();
            sumOfCurrLevel = 0;
            while (size-- > 0) {
                TreeNode head = qu.poll();
                sumOfCurrLevel += head.val;
                // if the left child of the head is not null
                if (head.left!= null) {
                    //push the child into the queue
                    qu.offer(head.left);
                }
                // if the right child is not null
                if (head.right!= null) {
                   // push the child into the queue
                    qu.offer(head.right);
                }
            }
        }
        return sumOfCurrLevel;
    }
    public static void main(String[] args)
    {
 
        TreeNode root;
        root = createNode(1);
        root.left = createNode(2);
        root.right = createNode(3);
        root.left.left = createNode(4);
        root.left.right = createNode(5);
        root.right.left = createNode(6);
        root.right.right = createNode(7);
        System.out.println(deepestLeavesSum(root));
    }
}
// this code is contributed by Rohan Raj
Output
22

 Time complexity: O(N) where N is the number of nodes in the tree.  

 

My Personal Notes arrow_drop_up
Recommended Articles
Page :