打印距离给定节点 K 处的所有节点：迭代方法

给定一棵二叉树、一个目标节点和一个整数K ，任务是找到与给定目标节点距离为K 的所有节点。

Consider the above-given Tree, For the target node 12. 
Input: K = 1 
Output: 8 10 14

Input: K = 2 
Output: 4 20

Input: K = 3 
output: 22 

方法：
距离K的节点一般有两种情况：

1. 距离 K 处的节点是目标节点的子节点。
2. 距离 K 处的节点是目标节点的祖先。

这个想法是在树上的层序遍历的帮助下将每个节点的父节点存储在一个哈希映射中。然后，在左子节点、右子节点和父节点上使用广度优先搜索从目标节点简单地遍历节点。在节点到目标节点的距离等于K 的任何时刻，然后打印队列中的所有节点。

下面是上述方法的实现：

C++

// C++ implementation to print all
// the nodes from the given target
// node iterative approach
 
#include 
 
using namespace std;
 
// Structure of the Node
struct Node {
    int val;
    Node *left, *right;
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
unordered_map um;
 
// Function to store all nodes
// parent in unordered_map
void storeParent(Node* root)
{
 
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    queue q;
    q.push(root);
     
    // Loop to iterate until the
    // queue is not empty
    while (!q.empty()) {
        Node* p = q.front();
        q.pop();
         
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root) {
            um[p] = NULL;
        }
         
        // if left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p->left) {
            um[p->left] = p;
            q.push(p->left);
        }
        if (p->right) {
            um[p->right] = p;
            q.push(p->right);
        }
    }
}
 
// Function to find the nodes
// at distance K from give node
void nodeAtDistK(Node* root,
           Node* target, int k)
{
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    unordered_set s;
    int dist = 0;
    queue q;
    q.push(target);
    s.insert(target);
     
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (!q.empty()) {
 
        // if distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k) {
            while (!q.empty()) {
                cout << q.front()->val << " ";
                q.pop();
            }
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.size();
        for (int i = 0; i < size; i++) {
            Node* p = q.front();
            q.pop();
 
            // if the left of node is not
            // visited yet then push it in
            // queue and insert it in set as well
            if (p->left &&
                s.find(p->left) == s.end()) {
                q.push(p->left);
                s.insert(p->left);
            }
 
            // if the right of node is not visited
            // yet then push it in queue
            // and insert it in set as well
            if (p->right &&
                s.find(p->right) == s.end()) {
                q.push(p->right);
                s.insert(p->right);
            }
 
            // if the parent of node is not visited
            // yet then push it in queue and
            // insert it in set as well
            if (um[p] && s.find(um[p]) == s.end()) {
                q.push(um[p]);
                s.insert(um[p]);
            }
        }
        dist++;
    }
}
 
// Function to create a newnode
Node* newnode(int val)
{
    Node* temp = new Node;
    temp->val = val;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver Code
int main()
{
    Node* root = newnode(20);
    root->left = newnode(8);
    root->right = newnode(22);
    root->right->left = newnode(5);
    root->right->right = newnode(8);
    root->left->left = newnode(4);
    root->left->left->left = newnode(25);
    root->left->right = newnode(12);
    root->left->right->left =
                   newnode(10);
    root->left->right->left->left =
                   newnode(15);
    root->left->right->left->right =
                   newnode(18);
    root->left->right->left->right->right =
                   newnode(23);
    root->left->right->right =
                   newnode(14);
    Node* target = root->left->right;
    storeParent(root);
    nodeAtDistK(root, target, 3);
    return 0;
}

Java

// Java implementation to print all
// the nodes from the given target
// node iterative approach
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
 
class GFG{
 
// Structure of the Node
static class Node
{
    int val;
    Node left, right;
 
    public Node(int val)
    {
        this.val = val;
        this.left = this.right = null;
    }
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
static HashMap um = new HashMap<>();
 
// Function to store all nodes
// parent in unordered_map
static void storeParent(Node root)
{
     
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    Queue q = new LinkedList<>();
    q.add(root);
 
    // Loop to iterate until the
    // queue is not empty
    while (!q.isEmpty())
    {
        Node p = q.poll();
 
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root)
        {
            um.put(p, null);
        }
 
        // if left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p.left != null)
        {
            um.put(p.left, p);
            q.add(p.left);
        }
        if (p.right != null)
        {
            um.put(p.right, p);
            q.add(p.right);
        }
    }
}
 
// Function to find the nodes
// at distance K from give node
static void nodeAtDistK(Node root,
                        Node target, int k)
{
     
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    HashSet s = new HashSet<>();
    int dist = 0;
     
    Queue q = new LinkedList<>();
    q.add(target);
    s.add(target);
 
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (!q.isEmpty())
    {
         
        // If distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k)
        {
            while (!q.isEmpty())
            {
                System.out.print(q.peek().val + " ");
                q.poll();
            }
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.size();
        for(int i = 0; i < size; i++)
        {
            Node p = q.poll();
 
            // If the left of node is not
            // visited yet then add it in
            // queue and insert it in set as well
            if (p.left != null && !s.contains(p.left))
            {
                q.add(p.left);
                s.add(p.left);
            }
 
            // If the right of node is not visited
            // yet then add it in queue
            // and insert it in set as well
            if (p.right != null && !s.contains(p.right))
            {
                q.add(p.right);
                s.add(p.right);
            }
 
            // If the parent of node is not visited
            // yet then add it in queue and
            // insert it in set as well
            if (um.get(p) != null &&
                !s.contains(um.get(p)))
            {
                q.add(um.get(p));
                s.add(um.get(p));
            }
        }
        dist++;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.right.left = new Node(5);
    root.right.right = new Node(8);
    root.left.left = new Node(4);
    root.left.left.left = new Node(25);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.left.left = new Node(15);
    root.left.right.left.right = new Node(18);
    root.left.right.left.right.right = new Node(23);
    root.left.right.right = new Node(14);
     
    Node target = root.left.right;
     
    storeParent(root);
     
    nodeAtDistK(root, target, 3);
}
}
 
// This code is contributed by sanjeev2552

C#

// C# implementation to print all
// the nodes from the given target
// node iterative approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Structure of the Node
public class Node
{
    public int val;
    public Node left, right;
 
    public Node(int val)
    {
        this.val = val;
        this.left = this.right = null;
    }
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
static Dictionary um = new Dictionary();
 
// Function to store all nodes
// parent in unordered_map
static void storeParent(Node root)
{
     
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    List q = new List();
    q.Add(root);
 
    // Loop to iterate until the
    // queue is not empty
    while (q.Count != 0)
    {
        Node p = q[0];
        q.RemoveAt(0);
         
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root)
        {
            um.Add(p, null);
        }
 
        // If left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p.left != null)
        {
            um.Add(p.left, p);
            q.Add(p.left);
        }
        if (p.right != null)
        {
            um.Add(p.right, p);
            q.Add(p.right);
        }
    }
}
 
// Function to find the nodes
// at distance K from give node
static void nodeAtDistK(Node root,
                        Node target, int k)
{
     
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    HashSet s = new HashSet();
    int dist = 0;
     
    List q = new List();
    q.Add(target);
    s.Add(target);
 
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (q.Count != 0)
    {
         
        // If distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k)
        {
            while (q.Count != 0)
            {
                Console.Write(q[0].val + " ");
                q.RemoveAt(0);
            }
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.Count;
        for(int i = 0; i < size; i++)
        {
            Node p = q[0];
            q.RemoveAt(0);
 
            // If the left of node is not
            // visited yet then add it in
            // queue and insert it in set as well
            if (p.left != null && !s.Contains(p.left))
            {
                q.Add(p.left);
                s.Add(p.left);
            }
 
            // If the right of node is not visited
            // yet then add it in queue and insert
            // it in set as well
            if (p.right != null && !s.Contains(p.right))
            {
                q.Add(p.right);
                s.Add(p.right);
            }
 
            // If the parent of node is not visited
            // yet then add it in queue and
            // insert it in set as well
            if (um[p] != null &&
                !s.Contains(um[p]))
            {
                q.Add(um[p]);
                s.Add(um[p]);
            }
        }
        dist++;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.right.left = new Node(5);
    root.right.right = new Node(8);
    root.left.left = new Node(4);
    root.left.left.left = new Node(25);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.left.left = new Node(15);
    root.left.right.left.right = new Node(18);
    root.left.right.left.right.right = new Node(23);
    root.left.right.right = new Node(14);
     
    Node target = root.left.right;
     
    storeParent(root);
     
    nodeAtDistK(root, target, 3);
}
}
 
// This code is contributed by Princi Singh

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