# Find the maximum path sum between two leaves of a binary tree

• Difficulty Level : Hard
• Last Updated : 23 Jun, 2022

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 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 root is empty, then 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.

1. Find maximum sum from leaf to root in left subtree of X (we can use this post for this and next steps)
2. Find maximum sum from leaf to root in right subtree of X.
3. 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.
4. 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.)

1. Maximum root to leaf path sum for the subtree rooted under current node.
2. 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 ` `using` `namespace` `std;`   `// A binary tree node` `struct` `Node` `{` `    ``int` `data;` `    ``struct` `Node* left, *right;` `};`   `// Utility function to allocate memory for a new node` `struct` `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 integers` `int` `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_MIN` `int` `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 Code` `int` `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 tree` `class` `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;` `      ``Node setTree(Node root){` `      `  `      ``Node 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` `    ``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(Node node)` `    ``{` `        ``Res res = ``new` `Res();` `        ``res.val = Integer.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` `    ``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(root));` `    ``}` `}`   `// This code is improved by Rahul Soni`

## Python3

 `# Python program to find maximumpath sum between two leaves` `# of a binary tree`   `INT_MIN ``=` `-``2``*``*``32`   `# A binary tree node`     `class` `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_MIN`     `def` `maxPathSumUtil(root, res):`   `    ``# Base Case` `    ``if` `root ``is` `None``:` `        ``return` `0`   `    ``# 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]` `    ``maxPathSumUtil(root, res)` `    ``return` `res[``0``]`     `# Driver program to test above function` `root ``=` `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

 ``

Output

`Max pathSum of the given binary tree is 27`

Time complexity: O(n) where n is the number of nodes
Auxiliary Space: O(1)

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