979. Distribute Coins in Binary Tree

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 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.

Solutions

Solution 1: DFS

We define a function $\textit{dfs(node)}$, which represents the coin overload in the subtree rooted at $\textit{node}$, i.e., the number of coins minus the number of nodes. If $\textit{dfs(node)}$ is positive, it means the subtree has more coins than nodes, and the excess coins need to be moved out of the subtree; if $\textit{dfs(node)}$ is negative, it means the subtree has fewer coins than nodes, and the shortfall needs to be moved into the subtree.

In the function $\textit{dfs(node)}$, we first traverse the left and right subtrees to obtain the coin overload $\textit{left}$ and $\textit{right}$ of the left and right subtrees, respectively. Then, the current number of moves needs to be increased by $|\textit{left}| + |\textit{right}|$, which means moving the coins from the left and right subtrees to the current node. After that, we return the coin overload of the entire subtree, which is $\textit{left} + \textit{right} + \textit{node.val} - 1$.

Finally, we return the number of moves.

The time complexity is $O(n)$, and the space complexity is $O(h)$. Here, $n$ and $h$ respectively represent the number of nodes and the height of the binary tree.

Python3JavaC++GoTypeScript

# 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:
        def dfs(root):
            if root is None:
                return 0
            left, right = dfs(root.left), dfs(root.right)
            nonlocal ans
            ans += abs(left) + abs(right)
            return left + right + root.val - 1

        ans = 0
        dfs(root)
        return ans

/**
 * 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 ans;

    public int distributeCoins(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = dfs(root.left);
        int right = dfs(root.right);
        ans += Math.abs(left) + Math.abs(right);
        return left + right + root.val - 1;
    }
}

/**
 * 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 ans = 0;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int left = dfs(root->left);
            int right = dfs(root->right);
            ans += abs(left) + abs(right);
            return left + right + root->val - 1;
        };
        dfs(root);
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func distributeCoins(root *TreeNode) (ans int) {
    var dfs func(*TreeNode) int
    dfs = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        left, right := dfs(root.Left), dfs(root.Right)
        ans += abs(left) + abs(right)
        return left + right + root.Val - 1
    }
    dfs(root)
    return
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function distributeCoins(root: TreeNode | null): number {
    let ans = 0;
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return 0;
        }
        const left = dfs(root.left);
        const right = dfs(root.right);
        ans += Math.abs(left) + Math.abs(right);
        return left + right + root.val - 1;
    };
    dfs(root);
    return ans;
}

979. Distribute Coins in Binary Tree

Description

Solutions

Solution 1: DFS

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