Question

979. Distribute Coins in Binary Tree

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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

# 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;
}