124. Binary Tree Maximum Path Sum

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:

1. Termination condition (when to terminate recursion)
2. Recursively process the left and right subtrees
3. 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.

Python3JavaC++GoTypeScriptRustJavaScriptC#

# 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

/**
 * 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);
    }
}

/**
 * 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;
    }
};

/**
 * 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
}