動機

post order很重要!!

Problem

Given a binary tree root, return the maximum sum of all keys of any sub-tree which is also a Binary Search Tree (BST).

Assume a BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than the node's key.
  • Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [1,4,3,2,4,2,5,Leetcode,null,null,null,null,null,4,6]Output: 20Explanation: Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.

Example 2:

Input: root = [4,3,null,1,2]Output: 2Explanation: Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.

Example 3:

Input: root = [-4,-2,-5]Output: 0Explanation: All values are negatives. Return an empty BST.

Example 4:

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

Example 5:

Input: root = [5,4,8,3,null,6,3]Output: 7

 

Constraints:

  • The number of nodes in the tree is in the range [1, 4 * 104].
  • -4 * 104 <= Node.val <= 4 * 104

Sol

這裡要處理

  • 加總
  • 驗證BST

如果從上到下，會很痛苦；所以要想起從底部往上也是tree，可以一步一步往上加與驗證

class Solution:
    def maxSumBST(self, root: TreeNode) -> int:
        ret = 0
        def post(root):
            nonlocal ret
            if not root:
                return [float('inf'), float('-inf'), 0]
            else:
                lmin, lmax, lsum = post(root.left)
                rmin, rmax, rsum = post(root.right)
                if lsum is not None and rsum is not None and lmax < root.val < rmin:
                    ret = max(ret, root.val+lsum+rsum)
                    return [min(lmin,root.val), max(root.val, rmax), root.val+lsum+rsum]
                else:
                    return [0,0,None]
                
        post(root)
        return ret