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;
    }
};




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