Binary Tree Maximum Path Sum LeetCode Solution

Difficulty Level Hard
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Problem Statement

Binary Tree Maximum Path Sum LeetCode Solution – 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

Test Case 1:

Input:

root = [-10, 9, 20, null, null, 15, 7]

Binary Tree Maximum Path Sum LeetCode Solution

Output:

42

Explanation

The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42 .

Binary Tree Maximum Path Sum LeetCode Solution

Approach:

The idea behind this problem is the following:

  • For every node of the tree, find the max among : leftSum, rightSum, root.val, leftSum + root.val , rightSum + root.val, leftSum + rightSum + root.val
  • Save the max to the global max
  • Now, you can’t return the global max. You need to return the max of leftSum + root.val, rightSum + root.val or root.val. This ensures that you are looking at a contiguous series of nodes.

The idea is also DFS. From bottom to up, at every node, we have four choices:

  1. the left sub-path ==> node ==> the right sub-path.
  2. the left sub-path ==> node ==> upper. The left sub-path may yield a negative sum, in which case we set node->left sub-path to zero.
  3. the right sub-path ==> node ==>upper. The right sub-path may yield a negative sum, in which case we set node->right sub-path to zero.
  4. 0 ==> upper, which means we abandon the entire tree rooted at this node because of a negative-sum.

Noted: Negative node values are possible.

Code for Binary Tree Maximum Path Sum

C++ Program

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int maxPathSum(TreeNode *root, int &ma)
    {
        if (!root)
            return 0;
        int left =  maxPathSum(root->left, ma);
        int right = maxPathSum(root->right, ma);
        ma = max(ma, root->val + left + right);
        return max(max(0, max(left, right)) + root->val, 0);
    }

public:
    int maxPathSum(TreeNode* root)
    {
        if (!root)
            return INT_MIN;  //This INT_MIN is just to comply with the judge.
        
        int ma = root->val;
        
        maxPathSum(root, ma);
        
        return ma;
    }
};

Java Program

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int maxSum = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        dfs(root);
        return maxSum;
    }
    
    public int dfs(TreeNode node) {
        if (node == null) return 0;
        
        int left = Math.max(dfs(node.left), 0);
        int right = Math.max(dfs(node.right), 0);
        
        maxSum = Math.max(maxSum, node.val + left + right);
        return node.val + Math.max(left, right);
    }
}

Complexity Analysis for Binary Tree Maximum Path Sum LeetCode Solution

Time Complexity: O(N)

Space Complexity: O(H)

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