最小化二叉树一分为二后形成的子树之和的绝对差
给定一棵由N个节点组成的二叉树,任务是通过删除一条边将二叉树分成两个子树,以使子树和的绝对差最小化。
例子:
Input: 1
/ \
2 3
/ \ \
4 5 8
/ \
6 7
Output: 6
Explanation: Split the tree between vertex 3 and 8. Subtrees created have sums 21 and 15
Input: 1
/ \
2 3
/ \ / \
4 5 6 7
Output: 4
Explanation: Split the tree between vertex 1 and 3. Subtrees created have sums 16 and 12
方法:给定问题可以通过以下步骤解决:
- 将变量minDiff初始化为 Integer 的最大值,它将存储答案
- 使用后序遍历将当前节点、左子树和右子树之和存储在当前节点中
- 使用前序遍历并在每次递归调用时找到子树的总和,如果拆分介于:
- 当前节点和左子节点
- 当前节点和右子节点
- 如果在任何递归调用中子树的差异小于minDiff ,则更新minDiff
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
struct Node {
Node* left;
Node* right;
int val;
Node(int val)
{
this->val = val;
this->left = nullptr;
this->right = nullptr;
}
};
int postOrder(Node* root)
{
if (root == nullptr)
return 0;
root->val += postOrder(root->left) + postOrder(root->right);
return root->val;
}
void preOrder(Node* root, int minDiff[])
{
if (root == nullptr)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff = abs(root->left->val - (root->val - root->left->val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff = abs(root->right->val - (root->val - root->right->val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0] = min(minDiff[0], min(leftDiff, rightDiff));
return;
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
int minAbsDiff(Node* root)
{
// Reference variable
// to store the answer
int minDiff[1];
minDiff[0] = INT_MAX;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
int main()
{
// Construct the tree
Node* root = new Node(1);
root->left = new Node(2);
root->right = new Node(3);
root->left->left = new Node(4);
root->left->right = new Node(5);
root->right->left = new Node(6);
root->right->right = new Node(7);
// Print the output
cout << minAbsDiff(root);
return 0;
}
// This code is contributed by mukesh07.
Java
// Java implementation for the above approach
import java.lang.Math;
import java.io.*;
class GFG {
static class Node {
Node left, right;
int val;
public Node(int val)
{
this.val = val;
this.left = this.right = null;
}
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
public static int minAbsDiff(Node root)
{
// Reference variable
// to store the answer
int[] minDiff = new int[1];
minDiff[0] = Integer.MAX_VALUE;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
public static int postOrder(Node root)
{
if (root == null)
return 0;
root.val += postOrder(root.left)
+ postOrder(root.right);
return root.val;
}
public static void preOrder(Node root,
int[] minDiff)
{
if (root == null)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff
= Math.abs(root.left.val
- (root.val - root.left.val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff
= Math.abs(root.right.val
- (root.val - root.right.val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0]
= Math.min(minDiff[0],
Math.min(leftDiff, rightDiff));
return;
}
// Driver code
public static void main(String[] args)
{
// Construct the tree
Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
// Print the output
System.out.println(minAbsDiff(root));
}
}
Python3
# Python3 implementation for the above approach
import sys
# Structure of Node
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
# Function to calculate minimum absolute
# difference of subtrees after
# splitting the tree into two parts
def minAbsDiff(root):
# Reference variable
# to store the answer
minDiff = [0]*(1)
minDiff[0] = sys.maxsize
# Function to store sum of values
# of current node, left subtree and
# right subtree in place of
# current node's value
postOrder(root)
# Function to perform every
# possible split and calculate
# absolute difference of subtrees
preOrder(root, minDiff)
return minDiff[0]
def postOrder(root):
if (root == None):
return 0
root.val += postOrder(root.left) + postOrder(root.right)
return root.val
def preOrder(root, minDiff):
if (root == None):
return
# Absolute difference in subtrees
# if left edge is broken
leftDiff = abs(root.left.val - (root.val - root.left.val))
# Absolute difference in subtrees
# if right edge is broken
rightDiff = abs(root.right.val - (root.val - root.right.val))
# Update minDiff if a difference
# lesser than it is found
minDiff[0] = min(minDiff[0], min(leftDiff, rightDiff))
return
# Construct the tree
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
# Print the output
print(minAbsDiff(root))
# This code is contributed by rameshtravel07.
C#
// C# implementation for the above approach
using System;
public class GFG {
class Node {
public Node left, right;
public int val;
}
static Node newNode(int data)
{
Node newNode = new Node();
newNode.val = data;
newNode.left= null;
newNode.right = null;
return newNode;
}
// Function to calculate minimum absolute
// difference of subtrees after
// splitting the tree into two parts
static int minAbsDiff(Node root)
{
// Reference variable
// to store the answer
int[] minDiff = new int[1];
minDiff[0] = Int32.MaxValue;
// Function to store sum of values
// of current node, left subtree and
// right subtree in place of
// current node's value
postOrder(root);
// Function to perform every
// possible split and calculate
// absolute difference of subtrees
preOrder(root, minDiff);
return minDiff[0];
}
static int postOrder(Node root)
{
if (root == null)
return 0;
root.val += postOrder(root.left)
+ postOrder(root.right);
return root.val;
}
static void preOrder(Node root,
int[] minDiff)
{
if (root == null)
return;
// Absolute difference in subtrees
// if left edge is broken
int leftDiff
= Math.Abs(root.left.val
- (root.val - root.left.val));
// Absolute difference in subtrees
// if right edge is broken
int rightDiff
= Math.Abs(root.right.val
- (root.val - root.right.val));
// Update minDiff if a difference
// lesser than it is found
minDiff[0]
= Math.Min(minDiff[0],
Math.Min(leftDiff, rightDiff));
return;
}
// Driver code
public static void Main()
{
// Construct the tree
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
// Print the output
Console.Write(minAbsDiff(root));
}
}
// This code is contributed by SURENDRA_GANGWAR.
Javascript
输出:
4
时间复杂度:O(N)
辅助空间:O(1)