1021. Distribute Coins In Binary Tree¶

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1021. Distribute Coins in Binary Tree

Medium

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:

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.

Solution¶

/**
 * 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 res;
    public int distributeCoins(TreeNode root) {
        res = 0;
        dfs(root);
        return res;
    }
    private int dfs(TreeNode root) {
        if (root == null) return 0;

        int left = dfs(root.left);
        int right = dfs(root.right);

        res += Math.abs(left) + Math.abs(right);

        return (root.val - 1) + left + right;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here