Problem description:

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

Solution:

The dfs function returns the amount of coins each node need or have excessively. For each node, it will try to balance the amount of the coins used by its left child and right child.

And it will return a positive number if there is excessive coins which could be used by its parent node, or a negative number if current node or its children need coins.

Each coin (excessive or needed) need one step to be sent to the parent node.

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int res= 0;
    int distributeCoins(TreeNode* root) {
        dfs(root);
        return res;
    }
    
    int dfs(TreeNode* root){
        if(!root) return 0;
        int left= dfs(root->left), right= dfs(root->right);
        res += abs(left)+ abs(right);
        return root->val+ left+ right-1;
    }
};

time complexity: $O(n)$
space complexity: $O(logn)$
reference:
https://goo.gl/YkYRTs