LeetCode-in-All
124. Binary Tree Maximum Path Sum
Hard
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
Solution
// Definition for a binary tree node.
// 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut max = i32::MIN;
Solution::helper(&root, &mut max);
max
}
fn helper(node: &Option<Rc<RefCell<TreeNode>>>, max: &mut i32) -> i32 {
if let Some(n) = node {
let n = n.borrow();
let left = Solution::helper(&n.left, max).max(0);
let right = Solution::helper(&n.right, max).max(0);
let current = n.val + left + right;
*max = (*max).max(current);
return n.val + left.max(right);
}
0
}
}