Distribute Coins In Binary Tree Problem
Description
LeetCode Problem 979.
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 1:
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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:
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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.
Example 3:
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Input: root = [1,0,2]
Output: 2
Example 4:
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Input: root = [1,0,0,null,3]
Output: 4
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.
Sample C++ Code
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/**
* 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 res = 0;
int distributeCoins(TreeNode* root) {
traversal(root);
return res;
}
int traversal(TreeNode* root) {
if (root->left)
root->val += traversal(root->left);
if (root->right)
root->val += traversal(root->right);
int temp = root->val -1;
res += abs(temp);
return temp;
}
};