動機

十分有趣

Problem

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 path.

 

Example 1:

Input: root = [1,2,3]Output: 6Explanation: 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: 42Explanation: 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

Sol

總和有

  1. 只有左邊
  2. 只有右邊
  3. 只有根結點
  4. 左邊+根結點
  5. 右邊+根結點
  6. 左邊+右邊+根結點

如果父節點也想加上去只有3~5的可以把父節點的值也加上去 所以分成兩個回傳值

def dfs(root):
    if not root:
        return [float('-inf'), float('-inf')]
    else:
        gLeft, lLeft = dfs(root.left)
        gRight, lRight = dfs(root.right)
        now_connectable = max(root.val, root.val+lLeft, root.val+lRight)
        now_global = max(root.val+lLeft+lRight, gLeft, gRight, now_connectable)
        return [now_global, now_connectable]
        
class Solution:
    def maxPathSum(self, root: TreeNode) -> int:
        return dfs(root)[0]