1038. Binary Search Tree to Greater Sum Tree

Problem 1038

https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/ Given the root of a binary search tree with distinct values, modify it so that every node has a new value equal to the sum of the values of the original tree that are greater than or equal to node.val.

As a reminder, a binary search tree is a tree that satisfies these constraints:

• 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: [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Solution

數字會從最大（right）的開始處理，將每一次的 node 數值紀錄並且累加，就會是接下來的 node 所應該加上去的數字。

List

class Solution(object):
    ListNode = []
    def bstToGst(self, root):
        self.ListNode = []
        self.dfs( root )
        SumNum = 0

        for node in self.ListNode:
            SumNum += node.val
            node.val = SumNum
        return root
    
    def dfs(self, node):
        if node:
            if node.right:
                self.dfs( node.right )
            self.ListNode.append( node )
            if node.left:
                self.dfs( node.left )

input_sumnum

class SolutionII(object):
    def bstToGst(self, root):
        self.bstToGst2( root, 0 )
        return root
    
    def bstToGst2( self, root, SumNum ):
        if root.right:
            SumNum = self.bstToGst2( root.right, SumNum )
        root.val += SumNum
        SumNum = root.val
        if root.left:
            SumNum = self.bstToGst2( root.left, SumNum )
        return SumNum

int val

class SolutionIII(object):
    val = 0
    def bstToGst(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root.right:
            self.bstToGst( root.right )
        root.val += self.val
        self.val = root.val
        if root.left:
            self.bstToGst(root.left)
        return root