979. Distribute Coins in Binary Tree

Given the root of a binary tree with N nodes, each node in the tree has node.val coins, and there are N coins total.

In one move, we may choose two adjacent nodes and move one coin from one node to another. (The move may be from parent to child, or from child to parent.)

Return the number of moves required to make every node have exactly one coin.

Example 1:

Input: [3,0,0]
Output: 2
Explanation: From the root of the tree, we move one coin to its left child, and one coin to its right child.

Example 2:

Input: [0,3,0]
Output: 3
Explanation: From the left child of the root, we move two coins to the root [taking two moves].  Then, we move one coin from the root of the tree to the right child.

Example 3:

Input: [1,0,2]
Output: 2

Example 4:

Input: [1,0,0,null,3]
Output: 4

Note:

1<= N <= 100
0 <= node.val <= N

思路：想着怎么递归就好了，直接递归最后的结果是不行，需要退后一步，构造一个helper(root)函数，用来求解root.left和root.right都OK的时候，root节点的值和需要的操作数

然后有2点需要注意：

1. helper中root==None的时候，是返回1,0

2. 直接helper(root)并不能使得root满足题意，只能说root.left, root.right满足


class Solution:
    def distributeCoins(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def helper(root):
            if not root: return 1,0
            sl,resl = helper(root.left)
            sr,resr = helper(root.right)
            res = resl+resr+abs(sl-1)+abs(sr-1)
            s = root.val-(1-sl)-(1-sr)
            return s,res
        dummy = TreeNode(1)
        dummy.left = root
        return helper(dummy)[1]

979. Distribute Coins in Binary Tree

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