LeetCode 124 Binary Tree Maximum Path Sum (C++, Python)

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Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
       1
      / \
     2   3
Return 6.
Solution C++


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/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    int dfsMaxPathSum(TreeNode* node, int& sum) {
        if (node == NULL) return 0;
        int left = 0;
        if (node->left) left = dfsMaxPathSum(node->left, sum);
        int right = 0;
        if (node->right) right = dfsMaxPathSum(node->right, sum);
        int cur = node->val;
        if (left > 0) {
            cur += left;
        }
        if (right > 0) {
            cur += right;
        }
        sum = max(cur, sum);
        int tmp = max(left, right);
        return tmp > 0? tmp + node->val: node->val; 
    }
public:
    int maxPathSum(TreeNode *root) {
        int sum = numeric_limits<int>::min();
        dfsMaxPathSum(root, sum);
        return sum;
    }
};
Solution Python


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# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def intMaxPathSum(self, node, ret):
        if not node: return 0
        cur = node.val
        left = self.intMaxPathSum(node.left, ret)
        right = self.intMaxPathSum(node.right, ret)
        if left > 0:
            cur += left
        if right > 0:
            cur += right
        ret[0] = max(ret[0], cur)
        tmp = max(left, right)
        return tmp + node.val if (tmp > 0) else node.val
        
    # @param root, a tree node
    # @return an integer
    def maxPathSum(self, root):
        ret = [- 2 ** 31]
        self.intMaxPathSum(root, ret)
        return ret[0]

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