Interview Classic 150 Problems P124 Maximum Path Sum in Binary Tree

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Timeline

Timeline

2025-10-28

init

DFS

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Problem:

P124 Maximum Path Sum in Binary Tree

https://leetcode.cn/problems/binary-tree-maximum-path-sum/description/?envType=study-plan-v2&envId=top-interview-150

Using DFS, the maximum value should be the maximum contribution of the left subtree plus the maximum contribution of the right subtree plus the value of the current node

// Definition for a binary tree node.
struct Solution;
#[derive(Debug, PartialEq, Eq)]
pub struct TreeNode {
    pub val: i32,
    pub left: Option<Rc<RefCell<TreeNode>>>,
    pub right: Option<Rc<RefCell<TreeNode>>>,
}

impl TreeNode {
    #[inline]
    pub fn new(val: i32) -> Self {
        TreeNode {
            val,
            left: None,
            right: None,
        }
    }
}

use std::cell::RefCell;
use std::cmp::max;
use std::rc::Rc;

impl Solution {
    pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut max_sum = i32::MIN;
        Self::dfs(&root, &mut max_sum);
        max_sum
    }
    fn dfs(node: &Option<Rc<RefCell<TreeNode>>>, max_sum: &mut i32) -> i32 {
        if let Some(n) = node {
            let n = n.borrow();

            // The maximum contribution of the left and right subtrees (if negative, better not to include)
            let left_gain = max(Self::dfs(&n.left, max_sum), 0);
            let right_gain = max(Self::dfs(&n.right, max_sum), 0);

            // The maximum path sum with the current node as the highest point
            let current_path_sum = n.val + left_gain + right_gain;

            // Update the global maximum path sum
            *max_sum = max(*max_sum, current_path_sum);

            // Return the maximum contribution of the current node to its parent (can only choose one side)
            return n.val + max(left_gain, right_gain);
        }
        0
    }
}

fn main() {
    println!("Hello, world!");
}

c++

#include <climits>

class Solution {
private:
    int maxSum = INT_MIN;

public:
    int maxGain(TreeNode* node) {
        if (node == nullptr) {
            return 0;
        }

        // Recursively calculate the maximum contribution of the left and right child nodes
        // Only when the maximum contribution value is greater than 0, the corresponding child node is selected
        int leftGain = std::max(maxGain(node->left), 0);
        int rightGain = std::max(maxGain(node->right), 0);

        // The maximum path sum of a node depends on the node's value and the maximum contribution values of its left and right child nodes
        int priceNewpath = node->val + leftGain + rightGain;

        // Update the answer
        maxSum = std::max(maxSum, priceNewpath);

        // Return the maximum contribution value of the node
        return node->val + std::max(leftGain, rightGain);
    }

    int maxPathSum(TreeNode* root) {
        maxGain(root);
        return maxSum;
    }
};

leetcode hot 100 rewrite

#include <algorithm>
#include <climits>

class Solution {
    private:
        int sum;
        int __maxPathSum(TreeNode *root)
        {
                if (root == nullptr)
                        return 0;
                int left_sum = __maxPathSum(root->left);
                int right_sum = __maxPathSum(root->right);
                sum = std::max({ left_sum + right_sum + root->val, root->val, root->val + left_sum,
                                 root->val + right_sum, sum });

                return std::max(std::max(left_sum, right_sum) + root->val, root->val);
        }
    public:
        int maxPathSum(TreeNode *root)
        {
                sum = INT_MIN;
                __maxPathSum(root);
                return sum;
        }
};