Distribute Coins in Binary Tree_one node to the door-优快云博客

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/106702280

本文探讨了在一个由N个节点组成的二叉树中，每个节点初始拥有的硬币数，如何通过最少的移动步骤使每个节点恰好拥有一个硬币。采用分治策略，递归地计算每个节点所需的移动次数，最终实现硬币的均匀分布。

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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.

思路：divide and conquer，注意返回的物理意义是什么，返回的是这个点还多余或者欠的coin数目，有正负号，但是用全局变量记录abs的值，也就是需要move in/out的次数。注意，如果current node有coin，那么不需要move，否则需要move in一个coin，这个表示就是：

root.val + leftstep + rightstep - 1; O(N);

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int distributeCoins(TreeNode root) {
        int[] res = {0};
        dfs(root, res);
        return res[0];
    }
    
    private int dfs(TreeNode root, int[] res) {
        if(root == null) {
            return 0;
        }
        int leftstep = dfs(root.left, res);
        int rightstep = dfs(root.right, res);
        res[0] += Math.abs(leftstep) + Math.abs(rightstep);
        return root.val + leftstep + rightstep - 1;
    }
}