979. Distribute Coins in Binary Tree.md

979. Distribute Coins in Binary Tree

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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.

Example 3:

Input: [1,0,2]
Output: 2

Example 4:

Input: [1,0,0,null,3]
Output: 4

Note:

• 1<= N <= 100
• 0 <= node.val <= N

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Solution

• Java

  • mine

    DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 38.8 MB, less than 46.43% of Java online submissions

    public int res;
    public int distributeCoins(TreeNode root) {
        dfs(root);
        return res;
    }
    
    public int dfs(TreeNode node) {
        if (node == null) {
            return 0;
        }
        //calculate the left and right need or left. 
        int left = dfs(node.left);
        int right = dfs(node.right);
        //add the move step
        res += Math.abs(left);
        res += Math.abs(right);
        return node.val - 1 + left + right;
    }
    

  • the most votes

    Recursive Runtime: 0 ms, faster than 100.00%, Memory Usage: 38.8 MB, less than 46.43% of Java online submissions

    int res = 0;
    public int distributeCoins(TreeNode root) {
        dfs(root);
        return res;
    }
    
    public int dfs(TreeNode root) {
        if (root == null) return 0;
        int left = dfs(root.left), right = dfs(root.right);
        res += Math.abs(left) + Math.abs(right);
        return root.val + left + right - 1;
    }