Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree.
Example:
Input: Root of below tree
1
/ \
2 3Output: 6
Input:
Binary tree
Output: 42
Explanation: Max path sum is represented using green color nodes in the above binary tree
Approach: To solve the problem follow the below idea:
For each node there can be four ways that the max path goes through the node:
- Node only
- Max path through Left Child + Node
- Max path through Right Child + Node
- Max path through Left Child + Node + Max path through Right Child
The idea is to keep track of four paths and pick up the max one in the end. An important thing to note is, that the root of every subtree needs to return the maximum path sum such that at most one child of the root is involved. This is needed for the parent function call. In the below code, this sum is stored in ‘max_single’ and returned by the recursive function.
Follow the given steps to solve the problem:
- If the root is NULL, return 0(Base Case)
- Call the recursive function to find the max sum for the left and the right subtree
- In a variable store the maximum of (root->data, maximum of (leftSum, rightSum) + root->data)
- In another variable store the maximum of previous step and root->data + leftSum + rightSum
- Return the maximum of the previous step
Below is the implementation of the above approach:
C++
// C/C++ program to find maximum path 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 allocate a new nodestruct Node* newNode(int data){ struct Node* newNode = new Node; newNode->data = data; newNode->left = newNode->right = NULL; return (newNode);}// This function returns overall maximum path sum in 'res'// And returns max path sum going through root.int findMaxUtil(Node* root, int& res){ // Base Case if (root == NULL) return 0; // l and r store maximum path sum going through left and // right child of root respectively int l = findMaxUtil(root->left, res); int r = findMaxUtil(root->right, res); // Max path for parent call of root. This path must // include at-most one child of root int max_single = max(max(l, r) + root->data, root->data); // Max Top represents the sum when the Node under // consideration is the root of the maxsum path and no // ancestors of root are there in max sum path int max_top = max(max_single, l + r + root->data); res = max(res, max_top); // Store the Maximum Result. return max_single;}// Returns maximum path sum in tree with given rootint findMaxSum(Node* root){ // Initialize result int res = INT_MIN; // Compute and return result findMaxUtil(root, res); return res;}// Driver codeint main(void){ struct Node* root = newNode(10); root->left = newNode(2); root->right = newNode(10); root->left->left = newNode(20); root->left->right = newNode(1); root->right->right = newNode(-25); root->right->right->left = newNode(3); root->right->right->right = newNode(4); // Function call cout << "Max path sum is " << findMaxSum(root); return 0;} |
Java
// Java program to find maximum path sum in Binary Tree/* Class containing left and right child of current node and key value*/class Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; }}// An object of Res is passed around so that the// same value can be used by multiple recursive calls.class Res { public int val;}class BinaryTree { // Root of the Binary Tree Node root; // This function returns overall maximum path sum in // 'res' And returns max path sum going through root. int findMaxUtil(Node node, Res res) { // Base Case if (node == null) return 0; // l and r store maximum path sum going through left // and right child of root respectively int l = findMaxUtil(node.left, res); int r = findMaxUtil(node.right, res); // Max path for parent call of root. This path must // include at-most one child of root int max_single = Math.max( Math.max(l, r) + node.data, node.data); // Max Top represents the sum when the Node under // consideration is the root of the maxsum path and // no ancestors of root are there in max sum path int max_top = Math.max(max_single, l + r + node.data); // Store the Maximum Result. res.val = Math.max(res.val, max_top); return max_single; } int findMaxSum() { return findMaxSum(root); } // Returns maximum path sum in tree with given root int findMaxSum(Node node) { // Initialize result // int res2 = Integer.MIN_VALUE; Res res = new Res(); res.val = Integer.MIN_VALUE; // Compute and return result findMaxUtil(node, res); return res.val; } /* Driver code */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(2); tree.root.right = new Node(10); tree.root.left.left = new Node(20); tree.root.left.right = new Node(1); tree.root.right.right = new Node(-25); tree.root.right.right.left = new Node(3); tree.root.right.right.right = new Node(4); // Function call System.out.println("maximum path sum is : " + tree.findMaxSum()); }} |
Python
# Python3 program to find maximum path sum in Binary Tree# A Binary Tree Nodeclass Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None# This function returns overall maximum path sum in 'res'# And returns max path sum going through rootdef findMaxUtil(root): # Base Case if root is None: return 0 # l and r store maximum path sum going through left # and right child of root respectively l = findMaxUtil(root.left) r = findMaxUtil(root.right) # Max path for parent call of root. This path # must include at most one child of root max_single = max(max(l, r) + root.data, root.data) # Max top represents the sum when the node under # consideration is the root of the maxSum path and # no ancestor of root are there in max sum path max_top = max(max_single, l+r + root.data) # Static variable to store the changes # Store the maximum result findMaxUtil.res = max(findMaxUtil.res, max_top) return max_single# Return maximum path sum in tree with given rootdef findMaxSum(root): # Initialize result findMaxUtil.res = float("-inf") # Compute and return result findMaxUtil(root) return findMaxUtil.res# Driver codeif __name__ == '__main__': root = Node(10) root.left = Node(2) root.right = Node(10) root.left.left = Node(20) root.left.right = Node(1) root.right.right = Node(-25) root.right.right.left = Node(3) root.right.right.right = Node(4) # Function call print "Max path sum is ", findMaxSum(root)# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program to find maximum// path sum in Binary Treeusing System;/* Class containing left andright child of currentnode and key value*/public class Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }}// An object of Res is passed// around so that the same value// can be used by multiple recursive calls.class Res { public int val;}public class BinaryTree { // Root of the Binary Tree Node root; // This function returns overall // maximum path sum in 'res' And // returns max path sum going through root. int findMaxUtil(Node node, Res res) { // Base Case if (node == null) return 0; // l and r store maximum path // sum going through left and // right child of root respectively int l = findMaxUtil(node.left, res); int r = findMaxUtil(node.right, res); // Max path for parent call of root. // This path must include // at-most one child of root int max_single = Math.Max( Math.Max(l, r) + node.data, node.data); // Max Top represents the sum // when the Node under // consideration is the root // of the maxsum path and no // ancestors of root are there // in max sum path int max_top = Math.Max(max_single, l + r + node.data); // Store the Maximum Result. res.val = Math.Max(res.val, max_top); return max_single; } int findMaxSum() { return findMaxSum(root); } // Returns maximum path // sum in tree with given root int findMaxSum(Node node) { // Initialize result // int res2 = int.MinValue; Res res = new Res(); res.val = int.MinValue; // Compute and return result findMaxUtil(node, res); return res.val; } /* Driver code */ public static void Main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(2); tree.root.right = new Node(10); tree.root.left.left = new Node(20); tree.root.left.right = new Node(1); tree.root.right.right = new Node(-25); tree.root.right.right.left = new Node(3); tree.root.right.right.right = new Node(4); // Function call Console.WriteLine("maximum path sum is : " + tree.findMaxSum()); }}// This code is contributed Rajput-Ji. |
Javascript
<script> // JavaScript program to find maximum // path sum in Binary Tree class Node { constructor(item) { this.left = null; this.right = null; this.data = item; } } let val; // Root of the Binary Tree let root; // This function returns overall maximum path sum in 'res' // And returns max path sum going through root. function findMaxUtil(node) { // Base Case if (node == null) return 0; // l and r store maximum path sum going through left and // right child of root respectively let l = findMaxUtil(node.left); let r = findMaxUtil(node.right); // Max path for parent call of root. This path must // include at-most one child of root let max_single = Math.max(Math.max(l, r) + node.data, node.data); // Max Top represents the sum when the Node under // consideration is the root of the maxsum path and no // ancestors of root are there in max sum path let max_top = Math.max(max_single, l + r + node.data); // Store the Maximum Result. val = Math.max(val, max_top); return max_single; } function findMaxsum() { return findMaxSum(root); } // Returns maximum path sum in tree with given root function findMaxSum(node) { // Initialize result // int res2 = Integer.MIN_VALUE; val = Number.MIN_VALUE; // Compute and return result findMaxUtil(node); return val; } root = new Node(10); root.left = new Node(2); root.right = new Node(10); root.left.left = new Node(20); root.left.right = new Node(1); root.right.right = new Node(-25); root.right.right.left = new Node(3); root.right.right.right = new Node(4); document.write("Max path sum is : " + findMaxsum()); </script> |
Output
Max path sum is 42
Time Complexity: O(N) where N is the number of nodes in the Binary Tree
Auxiliary Space: O(N)
Using Iterative DFS:
The basic idea behind the iterative DFS approach to finding the maximum path sum in a binary tree is to traverse the tree using a stack, maintaining the state of each node as we visit it.
Follow the steps to implement the approach:
- Initialize the max_sum variable to INT_MIN and create a stack to perform iterative DFS. Push the root node with a state of 0 onto the stack.
- Pop the top element from the stack and check if it is null. If it is null, skip the rest of the loop. Otherwise, check the state of the node.
- If the state is 0, push the node onto the stack with a state of 1 and push its left child onto the stack with a state of 0.
- If the state is 1, push the node onto the stack with a state of 2 and push its right child onto the stack with a state of 0.
- If the state is 2, calculate the maximum path sum that goes through the current node. To do this, calculate the maximum path sum that goes through the left child and the maximum path sum that goes through the right child. If either of these sums is negative, then we don’t include that child in the path. We add the maximum path sum that goes from the current node to one of its children, and the current node’s value, to get the maximum path sum that goes through the current node. We update the max_sum variable if the maximum path sum that goes through the current node is greater than the current value of max_sum.
- Update the node’s value to be the maximum path sum that goes from the node to one of its children. This step is important because the maximum path sum that goes through the node will either include both of its children or just one of its children.
- Repeat steps 2 to 6 until the stack is empty.
- Return the max_sum variable.
Below is the implementation of the above approach:
C++
// C++ code to implement iterative DFS approach#include <bits/stdc++.h>using namespace std;// Definition of a binary tree nodestruct Node { int val; Node *left, *right; Node(int val) : val(val) , left(nullptr) , right(nullptr) { }};int findMaxSum(Node* root){ int max_sum = INT_MIN; stack<pair<Node*, int> > s; s.push(make_pair(root, 0)); while (!s.empty()) { auto node = s.top().first; int state = s.top().second; s.pop(); if (node == nullptr) { continue; } if (state == 0) { // first visit to the node s.push(make_pair(node, 1)); s.push(make_pair(node->left, 0)); } else if (state == 1) { // second visit to the node s.push(make_pair(node, 2)); s.push(make_pair(node->right, 0)); } else { // third visit to the node int left_sum = (node->left != nullptr) ? node->left->val : 0; int right_sum = (node->right != nullptr) ? node->right->val : 0; max_sum = max(max_sum, node->val + max(0, left_sum) + max(0, right_sum)); int max_child_sum = max(left_sum, right_sum); node->val += max(0, max_child_sum); } } return max_sum;}// Driver Codeint main(){ // Constructing the tree given in the above example Node* root = new Node(10); root->left = new Node(2); root->right = new Node(-25); root->left->left = new Node(20); root->left->right = new Node(1); root->right->left = new Node(3); root->right->right = new Node(4); int max_sum = findMaxSum(root); cout << "Maximum Path Sum: " << max_sum << endl; return 0;}// This code is contributed by Veerendra_Singh_Rajpoot |
Java
import java.util.AbstractMap.SimpleEntry;import java.util.Stack;// Definition of a binary tree nodeclass Node { int val; Node left, right; Node(int val) { this.val = val; this.left = this.right = null; }}public class Main { static int findMaxSum(Node root) { int max_sum = Integer.MIN_VALUE; Stack<SimpleEntry<Node, Integer> > stack = new Stack<>(); stack.push(new SimpleEntry<>(root, 0)); while (!stack.empty()) { SimpleEntry<Node, Integer> pair = stack.pop(); Node node = pair.getKey(); int state = pair.getValue(); if (node == null) { continue; } if (state == 0) { // first visit to the node stack.push(new SimpleEntry<>(node, 1)); stack.push(new SimpleEntry<>(node.left, 0)); } else if (state == 1) { // second visit to the node stack.push(new SimpleEntry<>(node, 2)); stack.push( new SimpleEntry<>(node.right, 0)); } else { // third visit to the node int left_sum = (node.left != null) ? node.left.val : 0; int right_sum = (node.right != null) ? node.right.val : 0; max_sum = Math.max( max_sum, node.val + Math.max(0, left_sum) + Math.max(0, right_sum)); int max_child_sum = Math.max(left_sum, right_sum); node.val += Math.max(0, max_child_sum); } } return max_sum; } // Driver Code public static void main(String[] args) { // Constructing the tree given in the above example Node root = new Node(10); root.left = new Node(2); root.right = new Node(-25); root.left.left = new Node(20); root.left.right = new Node(1); root.right.left = new Node(3); root.right.right = new Node(4); int max_sum = findMaxSum(root); System.out.println("Maximum Path Sum: " + max_sum); }}// This code is contributed by Veerendra_Singh_Rajpoot |
Python
class TreeNode: def __init__(self, val): self.val = val self.left = None self.right = Nonedef find_max_sum(root): max_sum = float('-inf') stack = [] stack.append((root, 0)) while stack: node, state = stack.pop() if node is None: continue if state == 0: # First visit to the node stack.append((node, 1)) stack.append((node.left, 0)) elif state == 1: # Second visit to the node stack.append((node, 2)) stack.append((node.right, 0)) else: # Third visit to the node left_sum = node.left.val if node.left is not None else 0 right_sum = node.right.val if node.right is not None else 0 max_sum = max(max_sum, node.val + max(0, left_sum) + max(0, right_sum)) max_child_sum = max(left_sum, right_sum) node.val += max(0, max_child_sum) return max_sum# Constructing the tree given in the above exampleroot = TreeNode(10)root.left = TreeNode(2)root.right = TreeNode(-25)root.left.left = TreeNode(20)root.left.right = TreeNode(1)root.right.left = TreeNode(3)root.right.right = TreeNode(4)max_sum = find_max_sum(root)print("Maximum Path Sum:", max_sum) |
C#
using System;using System.Collections.Generic;// Definition of a binary tree nodeclass Node{ public int val; public Node left, right; public Node(int val) { this.val = val; this.left = this.right = null; }}public class GFG{ static int FindMaxSum(Node root) { int max_sum = int.MinValue; Stack<Tuple<Node, int>> stack = new Stack<Tuple<Node, int>>(); stack.Push(new Tuple<Node, int>(root, 0)); while (stack.Count > 0) { Tuple<Node, int> pair = stack.Pop(); Node node = pair.Item1; int state = pair.Item2; if (node == null) { continue; } if (state == 0) { // first visit to the node stack.Push(new Tuple<Node, int>(node, 1)); stack.Push(new Tuple<Node, int>(node.left, 0)); } else if (state == 1) { // second visit to the node stack.Push(new Tuple<Node, int>(node, 2)); stack.Push(new Tuple<Node, int>(node.right, 0)); } else { // third visit to the node int left_sum = (node.left != null) ? node.left.val : 0; int right_sum = (node.right != null) ? node.right.val : 0; max_sum = Math.Max(max_sum, node.val + Math.Max(0, left_sum) + Math.Max(0, right_sum)); int max_child_sum = Math.Max(left_sum, right_sum); node.val += Math.Max(0, max_child_sum); } } return max_sum; } // Driver Code public static void Main(string[] args) { // Constructing the tree given in the above example Node root = new Node(10); root.left = new Node(2); root.right = new Node(-25); root.left.left = new Node(20); root.left.right = new Node(1); root.right.left = new Node(3); root.right.right = new Node(4); int max_sum = FindMaxSum(root); Console.WriteLine("Maximum Path Sum: " + max_sum); }} |
Javascript
// JavaScript code for above approachclass Node { constructor(val) { this.val = val; this.left = this.right = null; }}function findMaxSum(root) { let max_sum = Number.MIN_SAFE_INTEGER; const stack = []; stack.push({ node: root, state: 0 }); while (stack.length > 0) { const { node, state } = stack.pop(); if (node === null) { continue; } if (state === 0) { // first visit to the node stack.push({ node: node, state: 1 }); stack.push({ node: node.left, state: 0 }); } else if (state === 1) { // second visit to the node stack.push({ node: node, state: 2 }); stack.push({ node: node.right, state: 0 }); } else { // third visit to the node const left_sum = (node.left !== null) ? node.left.val : 0; const right_sum = (node.right !== null) ? node.right.val : 0; max_sum = Math.max( max_sum, node.val + Math.max(0, left_sum) + Math.max(0, right_sum) ); const max_child_sum = Math.max(left_sum, right_sum); node.val += Math.max(0, max_child_sum); } } return max_sum;}// Constructing the tree given in the above exampleconst root = new Node(10);root.left = new Node(2);root.right = new Node(-25);root.left.left = new Node(20);root.left.right = new Node(1);root.right.left = new Node(3);root.right.right = new Node(4);const max_sum = findMaxSum(root);console.log("Maximum Path Sum: " + max_sum); |
Output
Maximum Path Sum: 32
Time Complexity: O(n) , The time complexity of this approach is O(n), where n is the number of nodes in the tree, since we need to visit each node once.
Space Complexity: O(h) , The space complexity is O(h), where h is the height of the tree, since we need to store at most h nodes in the stack
This article is contributed by Anmol Varshney (FB Profile: https://www.facebook.com/anmolvarshney695). Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above
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