LeetCode每日一题(124. Binary Tree Maximum Path Sum)

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

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早期的题目还是很淳朴的， 放到今天，能给这个题一个 medium 都算是高估它了。

如果一个 path 经过一个 node， 那只有两种情况:

1. path 只是路过， 从 node 的 left 或者 right 挑一条更优的向下延伸
2. 该 node 是该 path 的顶点， path 从 left 和 right 分别向下延伸

面对情况 1， 我们能给上层节点的最优解就是 max(left + val, right + val, val)， 面对情况 2， 我们能给答案贡献的最优解就是 max(left + right + val, left + val, right + val, val)

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use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn rc(root: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32) -> i32 {
        if let Some(node) = root {
            let val = node.borrow().val;
            let left = Solution::rc(node.borrow_mut().left.take(), ans);
            let right = Solution::rc(node.borrow_mut().right.take(), ans);
            *ans = (*ans).max(left + right + val);
            *ans = (*ans).max(left + val);
            *ans = (*ans).max(right + val);
            *ans = (*ans).max(val);
            return (left + val).max(right + val).max(val);
        }
        0
    }
    pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut ans = i32::MIN;
        Solution::rc(root, &mut ans);
        ans
    }
}