计算 N 叉树中给定 K 个节点的共同祖先的数量

给定一个 N 叉树根和一个K节点列表，任务是找到树中给定K节点的共同祖先的数量。

例子：

Input: root = 3  
                    /   \   
                 2      1
               / \    / | \   
             9  7  8  6 3
K = {7, 2, 9}
Output: 2 
Explanation: The common ancestors of the nodes 7, 9 and 2 are 2 and 3

Input: root = 2 
                       \
                        1
                         \
                          0—4
                        / | \
                      9  3 8
K = {9, 8, 3, 4, 0}
Output: 3

方法：给定的问题可以通过使用后序遍历来解决。这个想法是找到K个节点的最低共同祖先，然后增加其上方每个节点的祖先计数，直到到达根节点。可以按照以下步骤解决问题：

• 将所有列表节点添加到集合中
• 在树上应用后序遍历：
  • 找到最低的共同祖先，然后在每次递归调用时开始增加节点数的值
• 返回计算的答案

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)
        {
            children = new ArrayList<>();
            this.val = val;
        }
    }
 
    // Function to find the number
    // of common ancestors in a tree
    public static int numberOfAncestors(
        Node root,
        List nodes)
    {
 
        // Initialize a set
        Set set = new HashSet<>();
 
        // Iterate the list of nodes
        // and add them in a set
        for (Node curr : nodes) {
            set.add(curr);
        }
 
        // Find LCA and return
        // number of ancestors
        return CAcount(root, set)[1].val;
    }
 
    // Function to find LCA and
    // count number of ancestors
    public static Node[] CAcount(
        Node root, Set set)
    {
 
        // If the current node
        // is a desired node
        if (set.contains(root)) {
 
            Node[] res = new Node[2];
            res[0] = root;
            res[1] = new Node(1);
            return res;
        }
 
        // If leaf node then return null
        if (root.children.size() == 0) {
 
            return new Node[2];
        }
 
        // To count number of desired nodes
        // in the children branches
        int childCount = 0;
 
        // Initialize a node to return
        Node[] ans = new Node[2];
 
        // Iterate through all children
        for (Node child : root.children) {
 
            Node[] res = CAcount(child, set);
 
            // Increment child count if
            // desired node is found
            if (res[0] != null)
                childCount++;
 
            // If first desired node is found
            if (childCount == 1
                && ans[0] == null) {
 
                ans = res;
            }
            else if (childCount > 1) {
 
                ans[0] = root;
                ans[1] = new Node(1);
                return ans;
            }
        }
 
        // If LCA found below then increment
        // number of common ancestors
        if (ans[0] != null)
            ans[1].val++;
 
        // Return the answer
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Initialize the tree
        Node zero = new Node(0);
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(5);
        Node six = new Node(6);
        Node seven = new Node(7);
        zero.children.add(one);
        zero.children.add(two);
        zero.children.add(three);
        one.children.add(four);
        one.children.add(five);
        five.children.add(six);
        five.children.add(seven);
 
        // List of nodes whose
        // ancestors are to be found
        List nodes = new ArrayList<>();
        nodes.add(four);
        nodes.add(six);
        nodes.add(seven);
 
        // Call the function
        // and print the result
        System.out.println(
            numberOfAncestors(zero, nodes));
    }
}

C#

// C# implementation for the above approach
using System;
using System.Collections.Generic;
 
public class GFG {
 
    class Node {
 
        public List children;
        public int val;
 
        // constructor
        public Node(int val)
        {
            children = new List();
            this.val = val;
        }
    }
 
    // Function to find the number
    // of common ancestors in a tree
    static int numberOfAncestors(
        Node root,
        List nodes)
    {
 
        // Initialize a set
        HashSet set = new HashSet();
 
        // Iterate the list of nodes
        // and add them in a set
        foreach (Node curr in nodes) {
            set.Add(curr);
        }
 
        // Find LCA and return
        // number of ancestors
        return CAcount(root, set)[1].val;
    }
 
    // Function to find LCA and
    // count number of ancestors
    static Node[] CAcount(
        Node root, HashSet set)
    {
 
        // If the current node
        // is a desired node
        if (set.Contains(root)) {
 
            Node[] res = new Node[2];
            res[0] = root;
            res[1] = new Node(1);
            return res;
        }
 
        // If leaf node then return null
        if (root.children.Count == 0) {
 
            return new Node[2];
        }
 
        // To count number of desired nodes
        // in the children branches
        int childCount = 0;
 
        // Initialize a node to return
        Node[] ans = new Node[2];
 
        // Iterate through all children
        foreach (Node child in root.children) {
 
            Node[] res = CAcount(child, set);
 
            // Increment child count if
            // desired node is found
            if (res[0] != null)
                childCount++;
 
            // If first desired node is found
            if (childCount == 1
                && ans[0] == null) {
 
                ans = res;
            }
            else if (childCount > 1) {
 
                ans[0] = root;
                ans[1] = new Node(1);
                return ans;
            }
        }
 
        // If LCA found below then increment
        // number of common ancestors
        if (ans[0] != null)
            ans[1].val++;
 
        // Return the answer
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        // Initialize the tree
        Node zero = new Node(0);
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(5);
        Node six = new Node(6);
        Node seven = new Node(7);
        zero.children.Add(one);
        zero.children.Add(two);
        zero.children.Add(three);
        one.children.Add(four);
        one.children.Add(five);
        five.children.Add(six);
        five.children.Add(seven);
 
        // List of nodes whose
        // ancestors are to be found
        List nodes = new List();
        nodes.Add(four);
        nodes.Add(six);
        nodes.Add(seven);
 
        // Call the function
        // and print the result
        Console.WriteLine(
            numberOfAncestors(zero, nodes));
    }
}
 
// This code is contributed by shikhasingrajput

Javascript

// Javascript implementation for the above approach
 
 
 
class Node {
  // constructor
  constructor(val) {
    this.children = new Array();
    this.val = val;
  }
}
 
// Function to find the number
// of common ancestors in a tree
function numberOfAncestors(root, nodes) {
 
  // Initialize a set
  let set = new Set();
 
  // Iterate the list of nodes
  // and add them in a set
  for (curr of nodes) {
    set.add(curr);
  }
 
  // Find LCA and return
  // number of ancestors
  return CAcount(root, set)[1].val;
}
 
// Function to find LCA and
// count number of ancestors
function CAcount(root, set) {
 
  // If the current node
  // is a desired node
  if (set.has(root)) {
 
    let res = new Node(2);
    res[0] = root;
    res[1] = new Node(1);
    return res;
  }
 
  // If leaf node then return null
  if (root.children.length == 0) {
 
    return new Node(2);
  }
 
  // To count number of desired nodes
  // in the children branches
  let childCount = 0;
 
  // Initialize a node to return
  let ans = new Node(2);
 
  // Iterate through all children
  for (child of root.children) {
 
    let res = CAcount(child, set);
 
    // Increment child count if
    // desired node is found
    if (res[0] != null)
      childCount++;
 
    // If first desired node is found
    if (childCount == 1
      && ans[0] == null) {
 
      ans = res;
    }
    else if (childCount > 1) {
 
      ans[0] = root;
      ans[1] = new Node(1);
      return ans;
    }
  }
 
  // If LCA found below then increment
  // number of common ancestors
  if (ans[0] != null)
    ans[1].val++;
 
  // Return the answer
  return ans;
}
 
// Driver code
 
// Initialize the tree
let zero = new Node(0);
let one = new Node(1);
let two = new Node(2);
let three = new Node(3);
let four = new Node(4);
let five = new Node(5);
let six = new Node(6);
let seven = new Node(7);
zero.children.push(one);
zero.children.push(two);
zero.children.push(three);
one.children.push(four);
one.children.push(five);
five.children.push(six);
five.children.push(seven);
 
// List of nodes whose
// ancestors are to be found
let nodes = new Array();
nodes.push(four);
nodes.push(six);
nodes.push(seven);
 
// Call the function
// and print the result
document.write(numberOfAncestors(zero, nodes));
 
// This code is contributed by saurabh_jaiswal.

输出

2

时间复杂度： O(N)，其中 N 是树中的节点数
辅助空间： O(H)，H是树的高度