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📜  给定二叉树所有层级中非叶节点的最大和

📅  最后修改于: 2022-05-13 01:57:16.887000             🧑  作者: Mango

给定二叉树所有层级中非叶节点的最大和

给定具有正节点和负节点的二叉树,任务是在给定二叉树的所有级别中找到非叶节点的最大和。

例子:

Input:
                        4
                      /   \
                     2    -5
                    / \   
                  -1   3 
Output: 4
Sum of all non-leaf nodes at 0th level is 4.
Sum of all non-leaf nodes at 1st level is 2.
Sum of all non-leaf nodes at 2nd level is 0.
Hence maximum sum is 4

Input:
                 1
               /   \
             2      3
           /  \      \
          4    5      8
                    /   \
                   6     7  
Output: 8

方法:解决上述问题的思路是对树进行层序遍历。在进行遍历时,分别处理不同层次的节点。对于正在处理的每个级别,计算该级别中非叶节点的总和并跟踪最大总和。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Function to return the maximum sum of non-leaf nodes
// at any level in tree using level order traversal
int maxNonLeafNodesSum(struct Node* root)
{
    // Base case
    if (root == NULL)
        return 0;
 
    // Initialize result
    int result = 0;
 
    // Do Level order traversal keeping track
    // of the number of nodes at every level
    queue q;
    q.push(root);
    while (!q.empty()) {
 
        // Get the size of queue when the level order
        // traversal for one level finishes
        int count = q.size();
 
        // Iterate for all the nodes in the queue currently
        int sum = 0;
        while (count--) {
 
            // Dequeue a node from queue
            Node* temp = q.front();
            q.pop();
 
            // Add non-leaf node's value to current sum
            if (temp->left != NULL || temp->right != NULL)
                sum = sum + temp->data;
 
            // Enqueue left and right children of
            // dequeued node
            if (temp->left != NULL)
                q.push(temp->left);
            if (temp->right != NULL)
                q.push(temp->right);
        }
 
        // Update the maximum sum of leaf nodes value
        result = max(sum, result);
    }
 
    return result;
}
 
// Helper function that allocates a new node with the
// given data and NULL left and right pointers
struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
 
// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(8);
    root->right->right->left = newNode(6);
    root->right->right->right = newNode(7);
    cout << maxNonLeafNodesSum(root) << endl;
 
    return 0;
}


Java
// Java implementation of
// the above approach
import java.util.LinkedList;
import java.util.Queue;
class GFG{
 
// A binary tree node has data,
// pointer to left child
// and a pointer to right child
static class Node
{
  int data;
  Node left, right;
  public Node(int data)
  {
    this.data = data;
    this.left = this.right = null;
  }
};
 
// Function to return the maximum
// sum of non-leaf nodes  at any
// level in tree using level
// order traversal
static int maxNonLeafNodesSum(Node root)
{
  // Base case
  if (root == null)
    return 0;
 
  // Initialize result
  int result = 0;
 
  // Do Level order traversal keeping track
  // of the number of nodes at every level
  Queue q = new LinkedList<>();
  q.add(root);
 
  while (!q.isEmpty())
  {
    // Get the size of queue
    // when the level order
    // traversal for one
    // level finishes
    int count = q.size();
 
    // Iterate for all the nodes
    // in the queue currently
    int sum = 0;
    while (count-- > 0)
    {
      // Dequeue a node
      // from queue
      Node temp = q.poll();
 
      // Add non-leaf node's
      // value to current sum
      if (temp.left != null ||
          temp.right != null)
        sum = sum + temp.data;
 
      // Enqueue left and right
      // children of dequeued node
      if (temp.left != null)
        q.add(temp.left);
      if (temp.right != null)
        q.add(temp.right);
    }
 
    // Update the maximum sum
    // of leaf nodes value
    result = max(sum, result);
  }
  return result;
}
 
static int max(int sum,
               int result)
{
  if (sum > result)
    return sum;
  return result;
}
 
// Driver code
public static void main(String[] args)
{
  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.right = new Node(8);
  root.right.right.left = new Node(6);
  root.right.right.right = new Node(7);
  System.out.println(maxNonLeafNodesSum(root));
}   
}
 
// This code is contributed by sanjeev2552


Python3
# Python3 implementation of the approach
import queue
 
# A binary tree node has data, pointer to
# left child and a pointer to right child
class Node:
 
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
         
# Function to return the maximum Sum of 
# non-leaf nodes at any level in tree
# using level order traversal
def maxNonLeafNodesSum(root):
 
    # Base case
    if root == None:
        return 0
 
    # Initialize result
    result = 0
 
    # Do Level order traversal keeping track
    # of the number of nodes at every level
    q = queue.Queue()
    q.put(root)
    while not q.empty():
 
        # Get the size of queue when the level
        # order traversal for one level finishes
        count = q.qsize()
 
        # Iterate for all the nodes
        # in the queue currently
        Sum = 0
        while count:
 
            # Dequeue a node from queue
            temp = q.get()
             
            # Add non-leaf node's value to current Sum
            if temp.left != None or temp.right != None:
                Sum += temp.data
 
            # Enqueue left and right
            # children of dequeued node
            if temp.left != None:
                q.put(temp.left)
            if temp.right != None:
                q.put(temp.right)
                 
            count -= 1
         
        # Update the maximum Sum of leaf nodes value
        result = max(Sum, result)
     
    return result
 
# Driver code
if __name__ == "__main__":
 
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.right = Node(8)
    root.right.right.left = Node(6)
    root.right.right.right = Node(7)
    print(maxNonLeafNodesSum(root))
 
# This code is contributed by Rituraj Jain


C#
// C# implementation of
// the above approach
using System;
using System.Collections;
 
class GFG{
  
// A binary tree node has data,
// pointer to left child
// and a pointer to right child
class Node
{
  public int data;
  public Node left, right;
   
  public Node(int data)
  {
    this.data = data;
    this.left = this.right = null;
  }
};
  
// Function to return the maximum
// sum of non-leaf nodes  at any
// level in tree using level
// order traversal
static int maxNonLeafNodesSum(Node root)
{
   
  // Base case
  if (root == null)
    return 0;
  
  // Initialize result
  int result = 0;
  
  // Do Level order traversal keeping track
  // of the number of nodes at every level
  Queue q = new Queue();
  
  q.Enqueue(root);
  
  while (q.Count != 0)
  {
     
    // Get the size of queue
    // when the level order
    // traversal for one
    // level finishes
    int count = q.Count;
  
    // Iterate for all the nodes
    // in the queue currently
    int sum = 0;
     
    while (count-- > 0)
    {
       
      // Dequeue a node
      // from queue
      Node temp = (Node)q.Dequeue();
  
      // Add non-leaf node's
      // value to current sum
      if (temp.left != null ||
          temp.right != null)
        sum = sum + temp.data;
  
      // Enqueue left and right
      // children of dequeued node
      if (temp.left != null)
        q.Enqueue(temp.left);
      if (temp.right != null)
        q.Enqueue(temp.right);
    }
  
    // Update the maximum sum
    // of leaf nodes value
    result = max(sum, result);
  }
  return result;
}
  
static int max(int sum,
               int result)
{
  if (sum > result)
    return sum;
   
  return result;
}
  
// Driver code
public static void Main(string[] args)
{
  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.right = new Node(8);
  root.right.right.left = new Node(6);
  root.right.right.right = new Node(7);
  Console.Write(maxNonLeafNodesSum(root));
}   
}
 
// This code is contributed by rutvik_56


Javascript


输出:
8