leetcode 1038. Binary Search Tree to Greater Sum Tree 解法 python

一.问题描述

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]

 

Note:

1. The number of nodes in the tree is between 1 and 100.
2. Each node will have value between 0 and 100.
3. The given tree is a binary search tree.

二.解题思路

我的解题思路是：

首先计算出大于或等于原根节点的值，然后我们可以发现，对此根节点来说，它右边的部分的节点计算只需要减去它的父节点的节点值，然后如果有左孩子，再减去左孩子的和就是大于或等于它的值。

对于原根节点的左部分来说（注意是原根节点），每个节点的更新是父节点的值加上自己根节点的值，如果有右孩子，还要加上右孩子部分的和。

具体的更新看代码可能更明白一些。

更新：一个更明了的方法

对每个节点来说，大于或等于他的值的和都是，它如果是左孩子，就是本身的值加上右子树的值加上父节点的值加上父节点右子树的值，如果节点是右孩子，就是它本身的值加上右节点的值。可以看到我们中序遍历到最右子节点，先计算节点的右分支，然后计算当前节点，然后再更新左孩子

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三.源码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def bstToGst(self, root: TreeNode) -> TreeNode:
        new_root=TreeNode(None)
        sum_num=root.val
        sum_num+=self.findEqualAndLarge(root.right)
        new_root.val=sum_num
        if root.left:
            tmp1=TreeNode(None)
            new_root.left=tmp1
            self.updateLeftTree(root.left,sum_num,tmp1)
        if root.right:
            tmp2=TreeNode(None)
            new_root.right=tmp2
            self.updateRightTree(root.right,sum_num-root.val,tmp2)
        return new_root
    
    
    def updateLeftTree(self,root,sum_num,new_root):
        sum_num+=root.val
        new_root.val=sum_num
        if root.left==None and root.right==None:
            return
        if root.right:
            new_root.val+=self.findEqualAndLarge(root.right)
            temp1=TreeNode(None)
            new_root.right=temp1
            self.updateLeftTree(root.right,sum_num-root.val,temp1)
        if root.left:
            temp2=TreeNode(None)
            new_root.left=temp2
            self.updateLeftTree(root.left,new_root.val,temp2)
    
    def updateRightTree(self,root,sum_num,new_root):
        new_root.val=sum_num
        if root.left==None and root.right==None:
            return
        if root.left:
            new_root.val-=self.findEqualAndLarge(root.left)
            temp1=TreeNode(None)
            new_root.left=temp1
            self.updateRightTree(root.left,sum_num,temp1)
        if root.right:
            temp2=TreeNode(None)
            new_root.right=temp2
            self.updateRightTree(root.right,new_root.val-root.val,temp2)
    
    def findEqualAndLarge(self,root):
        sum_num=root.val
        if root.left==None and root.right==None:
            return sum_num
        if root.left:
            sum_num+=self.findEqualAndLarge(root.left)
        if root.right:
            sum_num+=self.findEqualAndLarge(root.right)
        return sum_num
            

#version 2
n = 0
def bstToGst(self, root):
    """
    :type root: TreeNode
    :rtype: TreeNode
    """
    if root:
        self.bstToGst(root.right)
        root.val = self.n = self.n + root.val
        self.bstToGst(root.left)
    return root