Sum of nodes at maximum depth of a Binary Tree | Iterative Approach

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, the sum of these nodes = 22

```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: There exists a recursive approach to this problem. This can also be solved using level order traversal and map. The idea is to do a traversal using a queue and keep track of current level. A map has been used to store the sum of nodes at the current level. Once all nodes are visited and the traversal is done, the last element of the map will contain the sum at the maximum depth of the tree.

Below is the implementation of the above approach:

C++

 `// C++ program to calculate the sum of ` `// nodes at the maximum depth of a binary tree ` `#include ` `using` `namespace` `std; ` ` `  `struct` `node { ` `    ``int` `data; ` `    ``node *left, *right; ` `} * temp; ` ` `  `node* newNode(``int` `data) ` `{ ` `    ``temp = ``new` `node; ` `    ``temp->data = data; ` `    ``temp->left = temp->right = NULL; ` ` `  `    ``return` `temp; ` `} ` ` `  `// Function to return the sum ` `int` `SumAtMaxLevel(node* root) ` `{ ` `    ``// Map to store level wise sum. ` `    ``map<``int``, ``int``> mp; ` ` `  `    ``// Queue for performing Level Order Traversal. ` `    ``// First entry is the node and ` `    ``// second entry is the level of this node. ` `    ``queue > q; ` ` `  `    ``// Root has level 0. ` `    ``q.push({ root, 0 }); ` ` `  `    ``while` `(!q.empty()) { ` ` `  `        ``// Get the node from front of Queue. ` `        ``pair temp = q.front(); ` `        ``q.pop(); ` ` `  `        ``// Get the depth of current node. ` `        ``int` `depth = temp.second; ` ` `  `        ``// Add the value of this node in map. ` `        ``mp[depth] += (temp.first)->data; ` ` `  `        ``// Push children of this node, ` `        ``// with increasing the depth. ` `        ``if` `(temp.first->left) ` `            ``q.push({ temp.first->left, depth + 1 }); ` ` `  `        ``if` `(temp.first->right) ` `            ``q.push({ temp.first->right, depth + 1 }); ` `    ``} ` ` `  `    ``map<``int``, ``int``>::iterator it; ` ` `  `    ``// Get the max depth from map. ` `    ``it = mp.end(); ` ` `  `    ``// last element ` `    ``it--; ` ` `  `    ``// return the max Depth sum. ` `    ``return` `it->second; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->right->left = newNode(6); ` `    ``root->right->right = newNode(7); ` ` `  `    ``cout << SumAtMaxLevel(root) << endl; ` `    ``return` `0; ` `} `

Python3

# Python3 program to calculate the
# sum of nodes at the maximum depth
# of a binary tree

# Helper function that allocates a
# new node with the given data and
# None left and right poers.
class newNode:

# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None

# Function to return the sum
def SumAtMaxLevel(root):

# Map to store level wise sum.
mp = {}

# Queue for performing Level Order
# Traversal. First entry is the node
# and second entry is the level of
# this node.
q = []

# Root has level 0.
q.append([root, 0])

while (len(q)):

# Get the node from front
# of Queue.
temp = q[0]
q.pop(0)

# Get the depth of current node.
depth = temp[1]

# Add the value of this node in map.
if depth not in mp:
mp[depth] = 0
mp[depth] += (temp[0]).data

# append children of this node,
# with increasing the depth.
if (temp[0].left) :
q.append([temp[0].left,
depth + 1])

if (temp[0].right) :
q.append([temp[0].right,
depth + 1])

# return the max Depth sum.
return list(mp.values())[-1]

# Driver Code
if __name__ == ‘__main__’:

# Let us construct the Tree
# shown in the above figure
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)
root.right.left = newNode(6)
root.right.right = newNode(7)
print(SumAtMaxLevel(root))

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

Output:

```22
```

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