124. Binary Tree Maximum Path Sum

Problem Statement

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

• The number of nodes in the tree is in the range [1, 3 * 104].

• -1000 <= Node.val <= 1000

Intuition

Approach:

At each step: Take either left or right
or left + right + root(current)


And in the parent we return, Max(l,r) + root

Links

https://leetcode.com/problems/binary-tree-maximum-path-sum/description/

Video Links

https://leetcode.com/problems/binary-tree-maximum-path-sum/solutions/603072/c-solution-o-n-with-detailed-explanation/

Approach 1:

C++

class Solution {
public:
    int max_sum=INT_MIN;
    int max_gain(TreeNode* root)
    {
        if(!root)
            return 0;
        int l=max(max_gain(root->left),0);
        int r=max(max_gain(root->right),0);

        int new_price=root->val+l+r;
        max_sum=max(max_sum,new_price);

        return root->val+max(l,r);
    }
    int maxPathSum(TreeNode* root) {
        max_gain(root);
        
        return max_sum;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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