979 Distribute Coins in Binary Tree 在二叉树中分配硬币
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.
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.
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.
题目描述:
给定一个有 N 个结点的二叉树的根结点 root，树中的每个结点上都对应有 node.val 枚硬币，并且总共有 N 枚硬币。
在一次移动中，我们可以选择两个相邻的结点，然后将一枚硬币从其中一个结点移动到另一个结点。(移动可以是从父结点到子结点，或者从子结点移动到父结点。)。
返回使每个结点上只有一枚硬币所需的移动次数。
示例 :
示例 1：
输入：[3,0,0]
输出：2
解释：从树的根结点开始，我们将一枚硬币移到它的左子结点上，一枚硬币移到它的右子结点上。
示例 2：
输入：[0,3,0]
输出：3
解释：从根结点的左子结点开始，我们将两枚硬币移到根结点上 [移动两次]。然后，我们把一枚硬币从根结点移到右子结点上。
示例 3：
输入：[1,0,2]
输出：2
示例 4：
输入：[1,0,0,null,3]
输出：4
提示:
1<= N <= 100
0 <= node.val <= N
思路:
后序遍历
叶子结点需要返回 x - 1 个金币给自己的父节点
这个 x - 1 有可能是正数负数或者 0
如果 x - 1 = 0, 说明其自身刚好有 1 个金币不用分配
否则结果需要加上 abs(x - 1) 表示需要给父节点或者从父节点分配 x - 1 个金币
时间复杂度为 O(n), 空间复杂度为 O(h), 其中 h 为树的高度
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
int distributeCoins(TreeNode* root)
{
int result = 0;
dfs(root, result);
return result;
}
private:
int dfs(TreeNode* root, int& result)
{
int coins = root -> val + (root -> right ? dfs(root -> right, result) : 0) + (root -> left ? dfs(root -> left, result) : 0) - 1;
result += abs(coins);
return coins;
}
};
Java:
/**
* 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 {
private int result = 0;
public int distributeCoins(TreeNode root) {
dfs(root);
return result;
}
private int dfs(TreeNode root) {
if (root == null) return 0;
if (root.left != null) root.val += dfs(root.left);
if (root.right != null) root.val += dfs(root.right);
result += Math.abs(root.val - 1);
return root.val - 1;
}
}
Python:
# 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:
result = 0
def dfs(root: TreeNode) -> int:
nonlocal result
coins = root.val + (dfs(root.right) if root.right else 0) + (dfs(root.left) if root.left else 0) - 1
result += abs(coins)
return coins
dfs(root)
return result