Question

1339. Maximum Product of Splitted Binary Tree

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Description

Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.

Return _the maximum product of the sums of the two subtrees_. Since the answer may be too large, return it modulo 109 + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

 

Example 1:

Input: root = [1,2,3,4,5,6]
Output: 110
Explanation: Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)

Example 2:

Input: root = [1,null,2,3,4,null,null,5,6]
Output: 90
Explanation: Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)

 

Constraints:

• The number of nodes in the tree is in the range [2, 5 * 104].
• 1 <= Node.val <= 104

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 maxProduct(self, root: Optional[TreeNode]) -> int:
    def sum(root: Optional[TreeNode]) -> int:
      if root is None:
        return 0
      return root.val + sum(root.left) + sum(root.right)

    def dfs(root: Optional[TreeNode]) -> int:
      if root is None:
        return 0
      t = root.val + dfs(root.left) + dfs(root.right)
      nonlocal ans, s
      if t < s:
        ans = max(ans, t * (s - t))
      return t

    mod = 10**9 + 7
    s = sum(root)
    ans = 0
    dfs(root)
    return ans % mod

/**
 * 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 long ans;
  private long s;

  public int maxProduct(TreeNode root) {
    final int mod = (int) 1e9 + 7;
    s = sum(root);
    dfs(root);
    return (int) (ans % mod);
  }

  private long dfs(TreeNode root) {
    if (root == null) {
      return 0;
    }
    long t = root.val + dfs(root.left) + dfs(root.right);
    if (t < s) {
      ans = Math.max(ans, t * (s - t));
    }
    return t;
  }

  private long sum(TreeNode root) {
    if (root == null) {
      return 0;
    }
    return root.val + sum(root.left) + sum(root.right);
  }
}

/**
 * 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 maxProduct(TreeNode* root) {
    using ll = long long;
    ll ans = 0;
    const int mod = 1e9 + 7;

    function<ll(TreeNode*)> sum = [&](TreeNode* root) -> ll {
      if (!root) {
        return 0;
      }
      return root->val + sum(root->left) + sum(root->right);
    };

    ll s = sum(root);

    function<ll(TreeNode*)> dfs = [&](TreeNode* root) -> ll {
      if (!root) {
        return 0;
      }
      ll t = root->val + dfs(root->left) + dfs(root->right);
      if (t < s) {
        ans = max(ans, t * (s - t));
      }
      return t;
    };

    dfs(root);
    return ans % mod;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func maxProduct(root *TreeNode) (ans int) {
  const mod = 1e9 + 7
  var sum func(*TreeNode) int
  sum = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    return root.Val + sum(root.Left) + sum(root.Right)
  }
  s := sum(root)
  var dfs func(*TreeNode) int
  dfs = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    t := root.Val + dfs(root.Left) + dfs(root.Right)
    if t < s {
      ans = max(ans, t*(s-t))
    }
    return t
  }
  dfs(root)
  ans %= mod
  return
}

/**
 * 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 maxProduct(root: TreeNode | null): number {
  const sum = (root: TreeNode | null): number => {
    if (!root) {
      return 0;
    }
    return root.val + sum(root.left) + sum(root.right);
  };
  const s = sum(root);
  let ans = 0;
  const mod = 1e9 + 7;
  const dfs = (root: TreeNode | null): number => {
    if (!root) {
      return 0;
    }
    const t = root.val + dfs(root.left) + dfs(root.right);
    if (t < s) {
      ans = Math.max(ans, t * (s - t));
    }
    return t;
  };
  dfs(root);
  return ans % mod;
}