979. Distribute Coins in Binary Tree

1. Description

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.

2. Example

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.

3. 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.

4. Solutions

Depth-First Search

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int distributeCoins(TreeNode *root) {
        traverse_(root);
        return count_;
    }

private:
    int count_ = 0;

    int traverse_(TreeNode *root) {
        if (root == nullptr) {
            return 0;
        } else {
            int diff = root->val + traverse_(root->left) + traverse_(root->right) - 1;

            count_ += abs(diff);
            return diff;
        }
    }
};