Description#
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
Solutions#
Solution 1: Recursion#
When thinking about the classic routine of recursion problems in binary trees, we consider:
- Termination condition (when to terminate recursion)
- Recursively process the left and right subtrees
- Merge the calculation results of the left and right subtrees
For this problem, we design a function $dfs(root)$, which returns the maximum path sum of the binary tree with $root$ as the root node.
The execution logic of the function $dfs(root)$ is as follows:
If $root$ does not exist, then $dfs(root)$ returns $0$;
Otherwise, we recursively calculate the maximum path sum of the left and right subtrees of $root$, denoted as $left$ and $right$. If $left$ is less than $0$, then we set it to $0$, similarly, if $right$ is less than $0$, then we set it to $0$.
Then, we update the answer with $root.val + left + right$. Finally, the function returns $root.val + \max(left, right)$.
In the main function, we call $dfs(root)$ to get the maximum path sum of each node, and the maximum value among them is the answer.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.
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| # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxPathSum(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
left = max(0, dfs(root.left))
right = max(0, dfs(root.right))
nonlocal ans
ans = max(ans, root.val + left + right)
return root.val + max(left, right)
ans = -inf
dfs(root)
return ans
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| /**
* 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 {
private int ans = -1001;
public int maxPathSum(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = Math.max(0, dfs(root.left));
int right = Math.max(0, dfs(root.right));
ans = Math.max(ans, root.val + left + right);
return root.val + Math.max(left, right);
}
}
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| /**
* 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 {
public:
int maxPathSum(TreeNode* root) {
int ans = -1001;
function<int(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return 0;
}
int left = max(0, dfs(root->left));
int right = max(0, dfs(root->right));
ans = max(ans, left + right + root->val);
return root->val + max(left, right);
};
dfs(root);
return ans;
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func maxPathSum(root *TreeNode) int {
ans := -1001
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left := max(0, dfs(root.Left))
right := max(0, dfs(root.Right))
ans = max(ans, left+right+root.Val)
return max(left, right) + root.Val
}
dfs(root)
return ans
}
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| /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function maxPathSum(root: TreeNode | null): number {
let ans = -1001;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const left = Math.max(0, dfs(root.left));
const right = Math.max(0, dfs(root.right));
ans = Math.max(ans, left + right + root.val);
return Math.max(left, right) + root.val;
};
dfs(root);
return ans;
}
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| // Definition for a binary tree node.
// #[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::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = (0).max(Self::dfs(&node.left, res));
let right = (0).max(Self::dfs(&node.right, res));
*res = (node.val + left + right).max(*res);
node.val + left.max(right)
}
pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut res = -1000;
Self::dfs(&root, &mut res);
res
}
}
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| /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var maxPathSum = function (root) {
let ans = -1001;
const dfs = root => {
if (!root) {
return 0;
}
const left = Math.max(0, dfs(root.left));
const right = Math.max(0, dfs(root.right));
ans = Math.max(ans, left + right + root.val);
return Math.max(left, right) + root.val;
};
dfs(root);
return ans;
};
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
private int ans = -1001;
public int MaxPathSum(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = Math.Max(0, dfs(root.left));
int right = Math.Max(0, dfs(root.right));
ans = Math.Max(ans, left + right + root.val);
return root.val + Math.Max(left, right);
}
}
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