Given a binary tree in which each node element contains a number. Find the maximum possible sum from one leaf node to another.
The maximum sum path may or may not go through the root. For example, in the following binary tree, the maximum sum is 27(3 + 6 + 9 + 0 – 1 + 10). Expected time complexity is O(n). If one side of the root is empty, then the function should return minus infinite (INT_MIN in case of C/C++)
A simple solution is to traverse the tree and do following for every traversed node X.
- Find maximum sum from leaf to root in left subtree of X (we can use this post for this and next steps)
- Find maximum sum from leaf to root in right subtree of X.
- Add the above two calculated values and X->data and compare the sum with the maximum value obtained so far and update the maximum value.
- Return the maximum value.
The time complexity of above solution is O(n2)
We can find the maximum sum using single traversal of binary tree. The idea is to maintain two values in recursive calls
(Note: If the tree is right-most or left-most tree then first we have to adjust the tree such that both the right and left are not null. Left-most means if the right of super root of the tree is null and right-most tree means if left of super root of the tree is null.)
- Maximum root to leaf path sum for the subtree rooted under current node.
- The maximum path sum between leaves (desired output).
For every visited node X, we find the maximum root to leaf sum in left and right subtrees of X. We add the two values with X->data, and compare the sum with maximum path sum found so far.
Following is the implementation of the above O(n) solution.
C++
// C++ program to find maximum path //sum between two leaves of a binary tree#include <bits/stdc++.h>using namespace std;// A binary tree nodestruct Node{ int data; struct Node* left, *right;};// Utility function to allocate memory for a new nodestruct Node* newNode(int data){ struct Node* node = new(struct Node); node->data = data; node->left = node->right = NULL; return (node);}// Utility function to find maximum of two integersint max(int a, int b){ return (a >= b)? a: b; }// A utility function to find the maximum sum between any// two leaves.This function calculates two values:// 1) Maximum path sum between two leaves which is stored// in res.// 2) The maximum root to leaf path sum which is returned.// If one side of root is empty, then it returns INT_MINint maxPathSumUtil(struct Node *root, int &res){ // Base cases if (root==NULL) return 0; if (!root->left && !root->right) return root->data; // Find maximum sum in left and right subtree. Also // find maximum root to leaf sums in left and right // subtrees and store them in ls and rs int ls = maxPathSumUtil(root->left, res); int rs = maxPathSumUtil(root->right, res); // If both left and right children exist if (root->left && root->right) { // Update result if needed res = max(res, ls + rs + root->data); // Return maximum possible value for root being // on one side return max(ls, rs) + root->data; } // If any of the two children is empty, return // root sum for root being on one side return (!root->left)? rs + root->data: ls + root->data;}// The main function which returns sum of the maximum// sum path between two leaves. This function mainly// uses maxPathSumUtil()int maxPathSum(struct Node *root){ int res = INT_MIN; int val = maxPathSumUtil(root, res); //--- for test case --- // 7 // / \ // Null -3 // (case - 1) // value of res will be INT_MIN but the answer is 4 , which is returned by the // function maxPathSumUtil(). if(root->left && root->right) return res; return max(res, val);}// Driver Codeint main(){ struct Node *root = newNode(-15); root->left = newNode(5); root->right = newNode(6); root->left->left = newNode(-8); root->left->right = newNode(1); root->left->left->left = newNode(2); root->left->left->right = newNode(6); root->right->left = newNode(3); root->right->right = newNode(9); root->right->right->right= newNode(0); root->right->right->right->left= newNode(4); root->right->right->right->right= newNode(-1); root->right->right->right->right->left= newNode(10); cout << "Max pathSum of the given binary tree is " << maxPathSum(root); return 0;} |
Java
// Java program to find maximum path sum between two leaves// of a binary treeclass Node { int data; Node left, right; 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 { int val;}class BinaryTree { static Node root; // A utility function to find the maximum sum between any // two leaves.This function calculates two values: // 1) Maximum path sum between two leaves which is stored // in res. // 2) The maximum root to leaf path sum which is returned. int maxPathSumUtil(Node node, Res res) { // Base cases if (node == null) return 0; if (node.left == null && node.right == null) return node.data; // Find maximum sum in left and right subtree. Also // find maximum root to leaf sums in left and right // subtrees and store them in ls and rs int ls = maxPathSumUtil(node.left, res); int rs = maxPathSumUtil(node.right, res); // If both left and right children exist if (node.left != null && node.right != null) { // Update result if needed res.val = Math.max(res.val, ls + rs + node.data); // Return maximum possible value for root being // on one side return Math.max(ls, rs) + node.data; } // If any of the two children is empty, return // root sum for root being on one side return (node.left == null) ? rs + node.data : ls + node.data; } // The main function which returns sum of the maximum // sum path between two leaves. This function mainly // uses maxPathSumUtil() int maxPathSum() { Res res = new Res(); res.val = Integer.MIN_VALUE; int val = maxPathSumUtil(root, res); if (root.left != null && root.right != null) return res.val; else { //--- for test case --- // 7 // / \ // Null -3 // value of res will be INT_MIN but the answer is 4, // which is returned by the function maxPathSumUtil() return Math.max(res.val,val); } } //Driver program to test above functions public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(-15); tree.root.left = new Node(5); tree.root.right = new Node(6); tree.root.left.left = new Node(-8); tree.root.left.right = new Node(1); tree.root.left.left.left = new Node(2); tree.root.left.left.right = new Node(6); tree.root.right.left = new Node(3); tree.root.right.right = new Node(9); tree.root.right.right.right = new Node(0); tree.root.right.right.right.left = new Node(4); tree.root.right.right.right.right = new Node(-1); tree.root.right.right.right.right.left = new Node(10); System.out.println("Max pathSum of the given binary tree is " + tree.maxPathSum()); }}// This code is improved by Fabio Missagia |
Python3
# Python program to find maximumpath sum between two leaves# of a binary treeINT_MIN = -2**32# A binary tree nodeclass Node: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None# Utility function to find maximum sum between any# two leaves. This function calculates two values:# 1) Maximum path sum between two leaves which are stored# in res# 2) The maximum root to leaf path sum which is returned# If one side of root is empty, then it returns INT_MINdef maxPathSumUtil(root, res): # Base Case if root is None: return 0 # if root is leaf node we can return root.data if not root.left and not root.right: return root.data # Find maximumsum in left and right subtree. Also # find maximum root to leaf sums in left and right # subtrees ans store them in ls and rs ls = maxPathSumUtil(root.left, res) rs = maxPathSumUtil(root.right, res) # If both left and right children exist if root.left is not None and root.right is not None: # update result if needed res[0] = max(res[0], ls + rs + root.data) # Return maximum possible value for root being # on one side return max(ls, rs) + root.data # If any of the two children is empty, return # root sum for root being on one side if root.left is None: return rs + root.data else: return ls + root.data# The main function which returns sum of the maximum# sum path betwee ntwo leaves. THis function mainly# uses maxPathSumUtil()def maxPathSum(root): res = [INT_MIN] res1 = maxPathSumUtil(root, res) # we have to check if root.left is None or root.right is None # for ex:- 10 # / \ # None -5 # this will return INT_MIN but answer is 5 which is res1 if root.left and root.right: return res[0] return max(res[0], res1)# Driver program to test above functionroot = Node(-15)root.left = Node(5)root.right = Node(6)root.left.left = Node(-8)root.left.right = Node(1)root.left.left.left = Node(2)root.left.left.right = Node(6)root.right.left = Node(3)root.right.right = Node(9)root.right.right.right = Node(0)root.right.right.right.left = Node(4)root.right.right.right.right = Node(-1)root.right.right.right.right.left = Node(10)print("Max pathSum of the given binary tree is", maxPathSum(root))# This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
using System;// C# program to find maximum path sum between two leaves // of a binary tree 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. public class Res{ public int val;}public class BinaryTree{ public static Node root; // A utility function to find the maximum sum between any // two leaves.This function calculates two values: // 1) Maximum path sum between two leaves which is stored // in res. // 2) The maximum root to leaf path sum which is returned. // If one side of root is empty, then it returns INT_MIN public virtual int maxPathSumUtil(Node node, Res res) { // Base cases if (node == null) { return 0; } if (node.left == null && node.right == null) { return node.data; } // Find maximum sum in left and right subtree. Also // find maximum root to leaf sums in left and right // subtrees and store them in ls and rs int ls = maxPathSumUtil(node.left, res); int rs = maxPathSumUtil(node.right, res); // If both left and right children exist if (node.left != null && node.right != null) { // Update result if needed res.val = Math.Max(res.val, ls + rs + node.data); // Return maximum possible value for root being // on one side return Math.Max(ls, rs) + node.data; } // If any of the two children is empty, return // root sum for root being on one side return (node.left == null) ? rs + node.data : ls + node.data; } // The main function which returns sum of the maximum // sum path between two leaves. This function mainly // uses maxPathSumUtil() public virtual int maxPathSum(Node node) { Res res = new Res(); res.val = int.MinValue; maxPathSumUtil(root, res); return res.val; } //Driver program to test above functions public static void Main(string[] args) { BinaryTree tree = new BinaryTree(); BinaryTree.root = new Node(-15); BinaryTree.root.left = new Node(5); BinaryTree.root.right = new Node(6); BinaryTree.root.left.left = new Node(-8); BinaryTree.root.left.right = new Node(1); BinaryTree.root.left.left.left = new Node(2); BinaryTree.root.left.left.right = new Node(6); BinaryTree.root.right.left = new Node(3); BinaryTree.root.right.right = new Node(9); BinaryTree.root.right.right.right = new Node(0); BinaryTree.root.right.right.right.left = new Node(4); BinaryTree.root.right.right.right.right = new Node(-1); BinaryTree.root.right.right.right.right.left = new Node(10); Console.WriteLine("Max pathSum of the given binary tree is " + tree.maxPathSum(root)); }} // This code is contributed by Shrikant13 |
Javascript
<script>// javascript program to find maximum path sum between two leaves// of a binary treeclass Node { constructor(val) { this.data = val; this.left = null; this.right = null; }}// An object of Res is passed around so that the// same value can be used by multiple recursive calls.class Res { constructor(){ this.val = 0; }}var root;function setTree(root) {var temp = new Node(0); // if tree is left most if (root.right == null) { root.right = temp; } else { // if tree is right most root.left = temp; } return root; } // A utility function to find the maximum sum between any // two leaves.This function calculates two values: // 1) Maximum path sum between two leaves which is stored // in res. // 2) The maximum root to leaf path sum which is returned. // If one side of root is empty, then it returns INT_MIN function maxPathSumUtil(node, res) { // Base cases if (node == null) return 0; if (node.left == null && node.right == null) return node.data; // Find maximum sum in left and right subtree. Also // find maximum root to leaf sums in left and right // subtrees and store them in ls and rs var ls = maxPathSumUtil(node.left, res); var rs = maxPathSumUtil(node.right, res); // If both left and right children exist if (node.left != null && node.right != null) { // Update result if needed res.val = Math.max(res.val, ls + rs + node.data); // Return maximum possible value for root being // on one side return Math.max(ls, rs) + node.data; } // If any of the two children is empty, return // root sum for root being on one side return (node.left == null) ? rs + node.data : ls + node.data; } // The main function which returns sum of the maximum // sum path between two leaves. This function mainly // uses maxPathSumUtil() function maxPathSum(node) { var res = new Res(); res.val = Number.MIN_VALUE; if (root.left == null || root.right == null) { root = setTree(root); } // if tree is left most or right most // call setTree() method to adjust tree first maxPathSumUtil(root, res); return res.val; } // Driver program to test above functions var root = new Node(-15); root.left = new Node(5); root.right = new Node(6); root.left.left = new Node(-8); root.left.right = new Node(1); root.left.left.left = new Node(2); root.left.left.right = new Node(6); root.right.left = new Node(3); root.right.right = new Node(9); root.right.right.right = new Node(0); root.right.right.right.left = new Node(4); root.right.right.right.right = new Node(-1); root.right.right.right.right.left = new Node(10); document.write("Max pathSum of the given binary tree is " + maxPathSum(root));// This code is contributed by umadevi9616 </script> |
Output
Max pathSum of the given binary tree is 27
Time complexity: O(n) where n is the number of nodes
Auxiliary Space: O(n)
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