leetcode -- 979. Distribute Coins in Binary Tree

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本文探讨了一种解决二叉树中金币分配问题的算法，通过深度优先搜索(DFS)策略，计算每个节点所需的金币移动次数，以确保每个节点恰好拥有一枚金币。通过实例演示了如何使用该算法有效解决各种二叉树结构中的金币分配问题。

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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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/distribute-coins-in-binary-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路：

1. 如果树的叶子是0，则过载量为-1，需要从它的父亲节点移动一枚金币到这个叶子节点上；如果树的叶子金币数m>1，则过载量为m-1，需要将这个叶子节点中的 m-1 枚金币移动到别的地方去.总的来说，对于一个叶子节点，需要移动到它中或需要从它移动到它的父亲中的金币数量为过载量 = abs(num_coins - 1)

2. dfs(root,res)表示个节点所在的子树中金币的过载量，即子树中金币的数量减去这个子树中节点的数量，接着计算这个节点与它的子节点之间需要移动金币的数量为 abs(dfs(node.left)) + abs(dfs(node.right))，这个节点金币的过载量为node.val + dfs(node.left) + dfs(node.right) - 1

/**
 * 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 dist(TreeNode* root, int &res){
        int coins = root->val + (root->right?dist(root->right,res):0) + (root->left?dist(root->left,res):0)-1;
        res+=abs(coins);
        return coins;
    }
    int distributeCoins(TreeNode* root) {
        int res = 0;
        dist(root,res);
        return res;
       
    }
};