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979. Distribute Coins in Binary Tree

You are given the root of a binary tree with n nodes where each node in the tree has node.val coins. There are n coins in total throughout the whole tree.

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

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

Example 1:

Input: root = [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: root = [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.

Constraints:

• The number of nodes in the tree is n.
• 1 <= n <= 100
• 0 <= Node.val <= n
• The sum of all Node.val is n.

Solutions (Python)

1. Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def distributeCoins(self, root: Optional[TreeNode]) -> int:
        return self.dfs(root)[2]

    def dfs(self, root: Optional[TreeNode]) -> (int, int, int):
        if root is None:
            return (0, 0, 0)

        lnodes, lvals, lmoves = self.dfs(root.left)
        rnodes, rvals, rmoves = self.dfs(root.right)

        return (lnodes + rnodes + 1,
                lvals + rvals + root.val,
                lmoves + rmoves + abs(lnodes - lvals) + abs(rnodes - rvals))