# Find largest subtree sum in a tree

In this article, we are going to see **how we can find the largest subtree sum in the given tree?**

Submitted by Radib Kar, on September 12, 2020

**Prerequisite:** Transform to sum tree

In this article, we are going to see how we can find the largest subtree sum in a tree?

In the below example,

**The largest subtree sum is 11 for the subtree rooted by 13.**

One thing is if all nodes are positive valued then, of course, the entire tree will have the largest sum tree sum which is basically the sum of all the nodes.

But since there can be nodes with negative values too, that's why we can't take the entire subtree and need to check for all subtree sum. Of course, the brute force method is to find subtree sum rooted by each node and to take the maximum sum. But the complexity would be * O(n^2)*.

So the efficient approach is a recursive solution using postorder traversal which will solve in one pass. The algorithm is below:

We have a variable ** ans** which returns the ultimate result and initialized as INT_MIN.

The recursive function is like below:

FUNCTION maxSubtreeSumRecur(node, ans){ If node is NULL return 0; node value is transformed to the subtree sum rooted by the node recursively Update the ans with updated node value if that's greater than ans return updated node value; END FUNCTION

So it's actually a postorder traversal which return the subtree sum and we update ** ans** if needed.

#include <bits/stdc++.h> using namespace std; // tree node is defined class TreeNode { public: int val; TreeNode* left; TreeNode* right; TreeNode(int data) { val = data; left = NULL; right = NULL; } }; int maxSubtreeSumRecur(TreeNode* node, int& ans) { if (!node) return 0; node->val += maxSubtreeSumRecur(node->left, ans) + maxSubtreeSumRecur(node->right, ans); ans = max(ans, node->val); return node->val; } int maxSubtreeSum(TreeNode* root) { int ans = INT_MIN; if (!root) return 0; maxSubtreeSumRecur(root, ans); return ans; } int main() { //building the tree like the example TreeNode* root = new TreeNode(-10); root->left = new TreeNode(7); root->right = new TreeNode(13); root->left->left = new TreeNode(-4); root->left->right = new TreeNode(3); root->right->left = new TreeNode(5); root->right->right = new TreeNode(-7); cout << "Maximum subtree sum is: " << maxSubtreeSum(root) << endl; return 0; }

**Output:**

Maximum subtree sum is: 11

**Dry Run:**

Since it's a postorder traversal we will show the dry run in bottom-up fashion as the recursion will reach there first before returning something.

**Figure 1: root->val -4 and subtree sum both 0, so total sum -4**

**Figure 2: root->val 3 and subtree sum both 0, so total sum 3**

**Figure 3: root->val 7 and subtree sum -4+3, so total sum 6**

**Figure 4: root->val 5 and subtree sum 0, so total sum 5**

**Figure 5: root->val -7 and subtree sum 0, so total sum -7**

**Figure 6: root->val 13 and subtree sum 5+(-7), so total sum 11**

**Figure 7: root->val -10 and subtree sum 6+11, so total sum 7**

Thus the largest subtree sum is 11 which appeared in Figure 6

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