Sum of nodes at maximum depth of a Binary Tree

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 current level.If the max level and current level are same ,add the root data to current sum.if level is less than max_level, do nothing.

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// Code to find the sum of the nodes
// which are present at the maximum depth.
#include <bits/stdc++.h> 
#include <iostream>
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;
}

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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 above approach:

Java

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// 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));
    }
}

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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#

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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

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Output :

22

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



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