Given a binary tree, find the maximum vertical level sum in binary tree.
Examples:
Input :
3
/ \
4 6
/ \ / \
-1 -2 5 10
\
8
Output : 14
Vertical level having nodes 6 and 8 has maximum
vertical sum 14.
Input :
1
/ \
5 8
/ \ \
2 -6 3
\ /
-1 -4
\
9
Output : 4
A simple solution is to first find vertical level sum of each level starting from minimum vertical level to maximum vertical level. Finding sum of one vertical level takes O(n) time. In worst case time complexity of this solution is O(n^2).
An efficient solution is to do level order traversal of given binary tree and update vertical level sum of each level while doing the traversal. After finding vertical sum of each level find maximum vertical sum from these values.
Steps to solve this problem:
1. Check if root is equal to null than return 0.
2. Declare an unordered map versum of integer key and value.
3. Declare variables maxsum=INT_MIN and curlev.
4. Declare a queue pair q of node* and integer.
5. Push (root,0) in q.
6. While q is not empty:
*Root= q.front().first.
*Curlev=q.front().second.
*Versum[curlev]+=root->data.
*Check if root->left is not null than push (root->left,curlev-1) in q.
*Check if root->right is not null than push (root->right,curlev+1) in q.
7. Iterate through every pair in versum as it:
*Update maxsum as max(maxsum,it.second).
8. Return maxsum.
Below is the implementation of above approach:
C++
// C++ program to find maximum vertical// sum in binary tree.#include <bits/stdc++.h>using namespace std;// A Binary Tree Nodestruct Node { int data; struct Node *left, *right;};// A utility function to create a new// Binary Tree Nodestruct Node* newNode(int item){ struct Node* temp = (struct Node*)malloc(sizeof(struct Node)); temp->data = item; temp->left = temp->right = NULL; return temp;}// Function to find maximum vertical sum// in binary tree.int maxVerticalSum(Node* root){ if (root == NULL) { return 0; } // To store sum of each vertical level. unordered_map<int, int> verSum; // To store maximum vertical level sum. int maxSum = INT_MIN; // To store vertical level of current node. int currLev; // Queue to perform level order traversal. // Each element of queue is a pair of node // and its vertical level. queue<pair<Node*, int> > q; q.push({ root, 0 }); while (!q.empty()) { // Extract node at front of queue // and its vertical level. root = q.front().first; currLev = q.front().second; q.pop(); // Update vertical level sum of // vertical level to which // current node belongs to. verSum[currLev] += root->data; if (root->left) q.push({ root->left, currLev - 1 }); if (root->right) q.push({ root->right, currLev + 1 }); } // Find maximum vertical level sum. for (auto it : verSum) maxSum = max(maxSum, it.second); return maxSum;}// Driver Program to test above functionsint main(){ /* 3 / \ 4 6 / \ / \ -1 -2 5 10 \ 8 */ struct Node* root = newNode(3); root->left = newNode(4); root->right = newNode(6); root->left->left = newNode(-1); root->left->right = newNode(-2); root->right->left = newNode(5); root->right->right = newNode(10); root->right->left->right = newNode(8); cout << maxVerticalSum(root); return 0;} |
Java
// Java code for the above approachimport java.util.*;class Node { int data; Node left, right; Node(int item) { data = item; left = right = null; }}class BinaryTree { // Function to find maximum vertical sum // in binary tree. static int maxVerticalSum(Node root) { if (root == null) { return 0; } // To store sum of each vertical level. Map<Integer, Integer> verSum = new HashMap<>(); // To store maximum vertical level sum. int maxSum = Integer.MIN_VALUE; // To store vertical level of current node. int currLev; // Queue to perform level order traversal. // Each element of queue is a pair of node // and its vertical level. Queue<AbstractMap.SimpleEntry<Node, Integer>> q = new LinkedList<>(); q.add(new AbstractMap.SimpleEntry<>(root, 0)); while (!q.isEmpty()) { // Extract node at front of queue // and its vertical level. AbstractMap.SimpleEntry<Node, Integer> front = q.poll(); root = front.getKey(); currLev = front.getValue(); // Update vertical level sum of // vertical level to which // current node belongs to. verSum.put(currLev, verSum.getOrDefault(currLev, 0) + root.data); if (root.left != null) q.add(new AbstractMap.SimpleEntry<>(root.left, currLev - 1)); if (root.right != null) q.add(new AbstractMap.SimpleEntry<>(root.right, currLev + 1)); } // Find maximum vertical level sum. for (int levelSum : verSum.values()) maxSum = Math.max(maxSum, levelSum); return maxSum; } public static void main(String[] args) { /* 3 / \ 4 6 / \ / \ -1 -2 5 10 \ 8 */ Node root = new Node(3); root.left = new Node(4); root.right = new Node(6); root.left.left = new Node(-1); root.left.right = new Node(-2); root.right.left = new Node(5); root.right.right = new Node(10); root.right.left.right = new Node(8); System.out.println(maxVerticalSum(root)); }}// This code is contributed by lokeshpotta20. |
Python3
# Python3 program to find maximum# vertical sum in binary tree.from sys import maxsizefrom collections import dequeINT_MIN = -maxsizeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None# Function to find maximum vertical sum# in binary tree.def maxVerticalSum(root: Node) -> int: if (root is None): return 0 # To store sum of each vertical level. verSum = dict() # To store maximum vertical level sum. maxSum = INT_MIN # To store vertical level of current node. currLev = 0 # Queue to perform level order traversal. # Each element of queue is a pair of node # and its vertical level. q = deque() q.append([root, 0]) while (q): # Extract node at front of queue # and its vertical level. root = q[0][0] currLev = q[0][1] q.popleft() # Update vertical level sum of # vertical level to which # current node belongs to. if currLev not in verSum: verSum[currLev] = 0 verSum[currLev] += root.data if (root.left): q.append([root.left, currLev - 1]) if (root.right): q.append([root.right, currLev + 1]) # Find maximum vertical level sum. for it in verSum: maxSum = max([maxSum, verSum[it]]) return maxSum# Driver codeif __name__ == "__main__": ''' 3 / \ 4 6 / \ / \ -1 -2 5 10 \ 8 ''' root = Node(3) root.left = Node(4) root.right = Node(6) root.left.left = Node(-1) root.left.right = Node(-2) root.right.left = Node(5) root.right.right = Node(10) root.right.left.right = Node(8) print(maxVerticalSum(root))# This code is contributed by sanjeev2552 |
C#
// C# code for the above approachusing System;using System.Collections.Generic;class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}public class GFG { // Function to find maximum vertical sum in binary tree. static int maxVerticalSum(Node root) { if (root == null) { return 0; } // To store sum of each vertical level. Dictionary<int, int> verSum = new Dictionary<int, int>(); // To store maximum vertical level sum. int maxSum = int.MinValue; // To store vertical level of current node. int currLev; // Queue to perform level order traversal. Each // element of queue is a pair of node and its // vertical level. Queue<KeyValuePair<Node, int> > q = new Queue<KeyValuePair<Node, int> >(); q.Enqueue(new KeyValuePair<Node, int>(root, 0)); while (q.Count > 0) { // Extract node at front of queue and its // vertical level. KeyValuePair<Node, int> front = q.Dequeue(); root = front.Key; currLev = front.Value; // Update vertical level sum of vertical level // to which current node belongs to. if (verSum.ContainsKey(currLev)) { verSum[currLev] += root.data; } else { verSum[currLev] = root.data; } if (root.left != null) { q.Enqueue(new KeyValuePair<Node, int>( root.left, currLev - 1)); } if (root.right != null) { q.Enqueue(new KeyValuePair<Node, int>( root.right, currLev + 1)); } } // Find maximum vertical level sum. foreach(int levelSum in verSum.Values) { maxSum = Math.Max(maxSum, levelSum); } return maxSum; } static public void Main() { // Code /* 3 / \ 4 6 / \ / \ -1 -2 5 10 \ 8 */ Node root = new Node(3); root.left = new Node(4); root.right = new Node(6); root.left.left = new Node(-1); root.left.right = new Node(-2); root.right.left = new Node(5); root.right.right = new Node(10); root.right.left.right = new Node(8); Console.WriteLine(maxVerticalSum(root)); }}// This code is contributed by sankar. |
Javascript
<script>// Javascript program to find maximum// vertical sum in binary tree.// A Binary Tree Nodeclass Node{ constructor(item) { this.left = null; this.right = null; this.data = item; }}// A utility function to create a new// Binary Tree Nodefunction newNode(item){ let temp = new Node(item); return temp;}// Function to find maximum vertical sum// in binary tree.function maxVerticalSum(root){ if (root == null) { return 0; } // To store sum of each vertical level. let verSum = new Map(); // To store maximum vertical level sum. let maxSum = Number.MIN_VALUE; // To store vertical level of current node. let currLev; // Queue to perform level order traversal. // Each element of queue is a pair of node // and its vertical level. let q = []; q.push([ root, 0 ]); while (q.length > 0) { // Extract node at front of queue // and its vertical level. root = q[0][0]; currLev = q[0][1]; q.shift(); // Update vertical level sum of // vertical level to which // current node belongs to. if (verSum.has(currLev)) { verSum.set(currLev, verSum.get(currLev) + root.data); } else { verSum.set(currLev, root.data); } if (root.left) q.push([root.left, currLev - 1]); if (root.right) q.push([root.right, currLev + 1]); } // Find maximum vertical level sum. verSum.forEach((values, keys)=>{ maxSum = Math.max(maxSum, values); }) return maxSum;}// Driver code/* 3 / \ 4 6 / \ / \ -1 -2 5 10 \ 8 */let root = newNode(3);root.left = newNode(4);root.right = newNode(6);root.left.left = newNode(-1);root.left.right = newNode(-2);root.right.left = newNode(5);root.right.right = newNode(10);root.right.left.right = newNode(8);document.write(maxVerticalSum(root));// This code is contributed by divyesh072019</script> |
Output
14
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(n)
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