给定 N-ary 树中的节点数，使得它们的子树是二叉树

给定一个 N 叉树根，任务是找到节点的数量，使得它们的子树是二叉树。

例子：

Input: Tree in the image below
 

Output: 11
Explanation: The nodes such that there subtree is a binary tree are {2, 8, 10, 6, 7, 3, 1, 9, 5, 11, 12}.

Input: Tree in the image below
 

Output: 9

方法：给定的问题可以通过使用后序遍历来解决。这个想法是使用递归并检查当前节点是否最多包含 2 个子节点以及子节点是否是有效的二叉树。可以按照以下步骤解决问题：

• 在 N 叉树上应用后序遍历：
  • 将每个子节点的返回值相加，计算在该节点找到的二叉树的数量，并将其存储在sum中
  • 如果根最多有两个孩子是有效的二叉树，那么根也是二叉树，所以返回一对sum + 1和1表示一个有效的二叉树
  • 如果根有两个以上的子节点或任何一个子节点不是有效的二叉树，则返回sum和0的对，表示无效的二叉树
• 从后序遍历返回的对中返回第一个索引总和处的值。

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
class Node
{
public:
    vector children;
    int val;
 
    // constructor
    Node(int v)
    {
 
        val = v;
        children = {};
    }
};
// Post-order traversal to find
// depth of all branches of every
// node of the tree
vector postOrder(Node *root)
{
 
    // Initialize a variable sum to
    // count number of binary trees
    int sum = 0;
 
    // Integer to indicate if the tree
    // rooted at current root is a
    // valid binary tree
    int valid = 1;
 
    // Use recursion on all child nodes
    for (Node *child : root->children)
    {
 
        // Get the number of binary trees
        vector binTrees = postOrder(child);
 
        // If tree rooted at current child
        // is not a valid binary tree then
        // tree rooted at current root is
        // also not a valid binary tree
        if (binTrees[1] == 0)
            valid = 0;
 
        // If branches are unbalanced
        // then store -1 in height
        sum += binTrees[0];
    }
 
    // Children are valid binary trees
    // and the number of children
    // are less than 3
    if (valid == 1 && root->children.size() < 3)
    {
 
        // Root is also a valid binary tree
        sum++;
    }
 
    // Children are leaf nodes but number
    // of children are greater than 2
    else
        valid = 0;
 
    // Return the answer
    return {sum, valid};
}
 
// Function to find the number of
// binary trees in an N-ary tree
int binTreesGeneric(Node *root)
{
 
    // Base case
    if (root == NULL)
        return 0;
 
    // Apply post-order traversal on
    // the root and return the answer
    return postOrder(root)[0];
}
 
// Driver code
int main()
{
 
    // Initialize the graph
    Node *twenty = new Node(20);
    Node *seven = new Node(7);
    Node *seven2 = new Node(7);
    Node *five = new Node(5);
    Node *four = new Node(4);
    Node *nine = new Node(9);
    Node *one = new Node(1);
    Node *two = new Node(2);
    Node *six = new Node(6);
    Node *eight = new Node(8);
    Node *ten = new Node(10);
    Node *three = new Node(3);
    Node *mfour = new Node(11);
    Node *zero = new Node(12);
    three->children.push_back(mfour);
    three->children.push_back(zero);
    ten->children.push_back(three);
    two->children.push_back(six);
    two->children.push_back(seven2);
    four->children.push_back(nine);
    four->children.push_back(one);
    four->children.push_back(five);
    seven->children.push_back(ten);
    seven->children.push_back(two);
    seven->children.push_back(eight);
    seven->children.push_back(four);
    twenty->children.push_back(seven);
 
    // Call the function
    // and print the result
    cout << (binTreesGeneric(twenty));
}
 
// This code is contributed by Potta Lokesh

Java

// Java implementation for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    static class Node {
 
        List children;
        int val;
 
        // constructor
        public Node(int val)
        {
 
            this.val = val;
            children = new ArrayList<>();
        }
    }
 
    // Function to find the number of
    // binary trees in an N-ary tree
    public static int binTreesGeneric(Node root)
    {
 
        // Base case
        if (root == null)
            return 0;
 
        // Apply post-order traversal on
        // the root and return the answer
        return postOrder(root)[0];
    }
 
    // Post-order traversal to find
    // depth of all branches of every
    // node of the tree
    public static int[] postOrder(Node root)
    {
 
        // Initialize a variable sum to
        // count number of binary trees
        int sum = 0;
 
        // Integer to indicate if the tree
        // rooted at current root is a
        // valid binary tree
        int valid = 1;
 
        // Use recursion on all child nodes
        for (Node child : root.children) {
 
            // Get the number of binary trees
            int[] binTrees = postOrder(child);
 
            // If tree rooted at current child
            // is not a valid binary tree then
            // tree rooted at current root is
            // also not a valid binary tree
            if (binTrees[1] == 0)
                valid = 0;
 
            // If branches are unbalanced
            // then store -1 in height
            sum += binTrees[0];
        }
 
        // Children are valid binary trees
        // and the number of children
        // are less than 3
        if (valid == 1 && root.children.size() < 3) {
 
            // Root is also a valid binary tree
            sum++;
        }
 
        // Children are leaf nodes but number
        // of children are greater than 2
        else
            valid = 0;
 
        // Return the answer
        return new int[] { sum, valid };
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Initialize the graph
        Node twenty = new Node(20);
        Node seven = new Node(7);
        Node seven2 = new Node(7);
        Node five = new Node(5);
        Node four = new Node(4);
        Node nine = new Node(9);
        Node one = new Node(1);
        Node two = new Node(2);
        Node six = new Node(6);
        Node eight = new Node(8);
        Node ten = new Node(10);
        Node three = new Node(3);
        Node mfour = new Node(11);
        Node zero = new Node(12);
        three.children.add(mfour);
        three.children.add(zero);
        ten.children.add(three);
        two.children.add(six);
        two.children.add(seven2);
        four.children.add(nine);
        four.children.add(one);
        four.children.add(five);
        seven.children.add(ten);
        seven.children.add(two);
        seven.children.add(eight);
        seven.children.add(four);
        twenty.children.add(seven);
 
        // Call the function
        // and print the result
        System.out.println(
            binTreesGeneric(twenty));
    }
}

Python3

# Python code for the above approach
class Node:
 
  # constructor
  def __init__(self, v):
    self.val = v;
    self.children = [];
 
# Post-order traversal to find
# depth of all branches of every
# node of the tree
def postOrder(root):
 
  # Initialize a variable sum to
  # count number of binary trees
  sum = 0;
 
  # Integer to indicate if the tree
  # rooted at current root is a
  # valid binary tree
  valid = 1;
 
  # Use recursion on all child nodes
  for child in root.children:
 
    # Get the number of binary trees
    binTrees = postOrder(child);
 
    # If tree rooted at current child
    # is not a valid binary tree then
    # tree rooted at current root is
    # also not a valid binary tree
    if (binTrees[1] == 0):
      valid = 0;
 
    # If branches are unbalanced
    # then store -1 in height
    sum += binTrees[0];
   
 
  # Children are valid binary trees
  # and the number of children
  # are less than 3
  if (valid == 1 and len(root.children) < 3):
 
    # Root is also a valid binary tree
    sum += 1
   
 
  # Children are leaf nodes but number
  # of children are greater than 2
  else:
    valid = 0;
 
  # Return the answer
  return [ sum, valid ];
 
 
# Function to find the number of
# binary trees in an N-ary tree
def binTreesGeneric(root):
 
  # Base case
  if (root == None):
    return 0;
 
  # Apply post-order traversal on
  # the root and return the answer
  return postOrder(root)[0];
 
 
# Driver code
 
# Initialize the graph
twenty = Node(20);
seven = Node(7);
seven2 = Node(7);
five = Node(5);
four = Node(4);
nine = Node(9);
one = Node(1);
two = Node(2);
six = Node(6);
eight = Node(8);
ten = Node(10);
three = Node(3);
mfour = Node(11);
zero = Node(12);
three.children.append(mfour);
three.children.append(zero);
ten.children.append(three);
two.children.append(six);
two.children.append(seven2);
four.children.append(nine);
four.children.append(one);
four.children.append(five);
seven.children.append(ten);
seven.children.append(two);
seven.children.append(eight);
seven.children.append(four);
twenty.children.append(seven);
 
# Call the function
# and print the result
print((binTreesGeneric(twenty)));
 
# This code is contributed by gfgking

Javascript

// Javascript code for the above approach
class Node {
 
  // constructor
  constructor(v) {
 
    this.val = v;
    this.children = [];
  }
};
 
// Post-order traversal to find
// depth of all branches of every
// node of the tree
function postOrder(root) {
 
  // Initialize a variable sum to
  // count number of binary trees
  let sum = 0;
 
  // Integer to indicate if the tree
  // rooted at current root is a
  // valid binary tree
  let valid = 1;
 
  // Use recursion on all child nodes
  for (child of root.children) {
 
    // Get the number of binary trees
    let binTrees = postOrder(child);
 
    // If tree rooted at current child
    // is not a valid binary tree then
    // tree rooted at current root is
    // also not a valid binary tree
    if (binTrees[1] == 0)
      valid = 0;
 
    // If branches are unbalanced
    // then store -1 in height
    sum += binTrees[0];
  }
 
  // Children are valid binary trees
  // and the number of children
  // are less than 3
  if (valid == 1 && root.children.length < 3) {
 
    // Root is also a valid binary tree
    sum++;
  }
 
  // Children are leaf nodes but number
  // of children are greater than 2
  else
    valid = 0;
 
  // Return the answer
  return [ sum, valid ];
}
 
// Function to find the number of
// binary trees in an N-ary tree
function binTreesGeneric(root) {
 
  // Base case
  if (root == null)
    return 0;
 
  // Apply post-order traversal on
  // the root and return the answer
  return postOrder(root)[0];
}
 
// Driver code
 
// Initialize the graph
let twenty = new Node(20);
let seven = new Node(7);
let seven2 = new Node(7);
let five = new Node(5);
let four = new Node(4);
let nine = new Node(9);
let one = new Node(1);
let two = new Node(2);
let six = new Node(6);
let eight = new Node(8);
let ten = new Node(10);
let three = new Node(3);
let mfour = new Node(11);
let zero = new Node(12);
three.children.push(mfour);
three.children.push(zero);
ten.children.push(three);
two.children.push(six);
two.children.push(seven2);
four.children.push(nine);
four.children.push(one);
four.children.push(five);
seven.children.push(ten);
seven.children.push(two);
seven.children.push(eight);
seven.children.push(four);
twenty.children.push(seven);
 
// Call the function
// and print the result
document.write((binTreesGeneric(twenty)));
 
// This code is contributed by gfgking

输出

11

时间复杂度： O(N)
辅助空间： O(N)