下面是递归BFS的Scala 2.11.4实现。为了简洁起见，我牺牲了尾部调用优化，但是TCOd版本非常相似。参见@snv的帖子。

import scala.collection.immutable.Queue

object RecursiveBfs {
  def bfs[A](tree: Tree[A], target: A): Boolean = {
    bfs(Queue(tree), target)
  }

  private def bfs[A](forest: Queue[Tree[A]], target: A): Boolean = {
    forest.dequeueOption exists {
      case (E, tail) => bfs(tail, target)
      case (Node(value, _, _), _) if value == target => true
      case (Node(_, l, r), tail) => bfs(tail.enqueue(List(l, r)), target)
    }
  }

  sealed trait Tree[+A]
  case class Node[+A](data: A, left: Tree[A], right: Tree[A]) extends Tree[A]
  case object E extends Tree[Nothing]
}

下面使用Haskell对我来说似乎很自然。在树的各个层次上递归迭代(这里我将名字收集到一个大的有序字符串中，以显示树的路径):

data Node = Node {name :: String, children :: [Node]}
aTree = Node "r" [Node "c1" [Node "gc1" [Node "ggc1" []], Node "gc2" []] , Node "c2" [Node "gc3" []], Node "c3" [] ]
breadthFirstOrder x = levelRecurser [x]
    where levelRecurser level = if length level == 0
                                then ""
                                else concat [name node ++ " " | node <- level] ++ levelRecurser (concat [children node | node <- level])

我已经用c++做了一个程序，它是在联合和不联合图工作。

    #include <queue>
#include "iostream"
#include "vector"
#include "queue"

using namespace std;

struct Edge {
    int source,destination;
};

class Graph{
    int V;
    vector<vector<int>> adjList;
public:

    Graph(vector<Edge> edges,int V){
        this->V = V;
        adjList.resize(V);
        for(auto i : edges){
            adjList[i.source].push_back(i.destination);
            //     adjList[i.destination].push_back(i.source);
        }
    }
    void BFSRecursivelyJoinandDisjointtGraphUtil(vector<bool> &discovered, queue<int> &q);
    void BFSRecursivelyJointandDisjointGraph(int s);
    void printGraph();


};

void Graph :: printGraph()
{
    for (int i = 0; i < this->adjList.size(); i++)
    {
        cout << i << " -- ";
        for (int v : this->adjList[i])
            cout <<"->"<< v << " ";
        cout << endl;
    }
}


void Graph ::BFSRecursivelyJoinandDisjointtGraphUtil(vector<bool> &discovered, queue<int> &q) {
    if (q.empty())
        return;
    int v = q.front();
    q.pop();
    cout << v <<" ";
    for (int u : this->adjList[v])
    {
        if (!discovered[u])
        {
            discovered[u] = true;
            q.push(u);
        }
    }
    BFSRecursivelyJoinandDisjointtGraphUtil(discovered, q);

}

void Graph ::BFSRecursivelyJointandDisjointGraph(int s) {
    vector<bool> discovered(V, false);
    queue<int> q;

    for (int i = s; i < V; i++) {
        if (discovered[i] == false)
        {
            discovered[i] = true;
            q.push(i);
            BFSRecursivelyJoinandDisjointtGraphUtil(discovered, q);
        }
    }
}

int main()
{

    vector<Edge> edges =
            {
                    {0, 1}, {0, 2}, {1, 2}, {2, 0}, {2,3},{3,3}
            };

    int V = 4;
    Graph graph(edges, V);
 //   graph.printGraph();
    graph.BFSRecursivelyJointandDisjointGraph(2);
    cout << "\n";




    edges = {
            {0,4},{1,2},{1,3},{1,4},{2,3},{3,4}
    };

    Graph graph2(edges,5);

    graph2.BFSRecursivelyJointandDisjointGraph(0);
    return 0;
}

我认为这可以使用指针来完成，而不使用任何队列。

基本上我们在任何地方都维护两个指针，一个指向父结点，另一个指向待处理的子结点(链接列表指向所有已处理的子结点)

现在你只需分配子进程的指针&当父进程处理完成时，你只需让子进程成为父进程进行下一层的处理

以下是我的代码:

//Tree Node
struct Node {
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
    

/ / Algorightm:

    void LevelTraverse(Node* parent,Node* chidstart,Node* childend ){
        if(!parent && !chidstart) return;  // we processed everything
        
        if(!parent && chidstart){ //finished processing last level
            parent=chidstart;chidstart=childend=NULL; // assgin child to parent for processing next level
            LevelTraverse(parent,chidstart,childend);
        }else if(parent && !chidstart){ // This is new level first node tobe processed
            Node* temp=parent; parent=parent->next;
            if(temp->left) { childend=chidstart=temp->left; }
            if(chidstart){
                if(temp->right) { childend->next=temp->right; childend=temp->right; }
            }else{
                if(temp->right) { childend=chidstart=temp->right; }
            }
            LevelTraverse(parent,chidstart,childend);
        }else if(parent && chidstart){ //we are in mid of some level processing
            Node* temp=parent; parent=parent->next;
            if(temp->left) { childend->next=temp->left; childend=temp->left; }
            if(temp->right) { childend->next=temp->right; childend=temp->right; }
            LevelTraverse(parent,chidstart,childend);
        }
    }

//驱动代码:

    Node* connect(Node* root) {
        if(!root) return NULL;
        Node* parent; Node* childs, *childe; parent=childs=childe=NULL;
        parent=root;
        LevelTraverse(parent, childs, childe);
        return root;
    }

下面是简短的Scala解决方案:

  def bfs(nodes: List[Node]): List[Node] = {
    if (nodes.nonEmpty) {
      nodes ++ bfs(nodes.flatMap(_.children))
    } else {
      List.empty
    }
  }

使用返回值作为累加器的想法是很适合的。 可以在其他语言中以类似的方式实现，只需确保您的递归函数处理的节点列表。

测试代码清单(使用@marco测试树):

import org.scalatest.FlatSpec

import scala.collection.mutable

class Node(val value: Int) {

  private val _children: mutable.ArrayBuffer[Node] = mutable.ArrayBuffer.empty

  def add(child: Node): Unit = _children += child

  def children = _children.toList

  override def toString: String = s"$value"
}

class BfsTestScala extends FlatSpec {

  //            1
  //          / | \
  //        2   3   4
  //      / |       | \
  //    5   6       7  8
  //  / |           | \
  // 9  10         11  12
  def tree(): Node = {
    val root = new Node(1)
    root.add(new Node(2))
    root.add(new Node(3))
    root.add(new Node(4))
    root.children(0).add(new Node(5))
    root.children(0).add(new Node(6))
    root.children(2).add(new Node(7))
    root.children(2).add(new Node(8))
    root.children(0).children(0).add(new Node(9))
    root.children(0).children(0).add(new Node(10))
    root.children(2).children(0).add(new Node(11))
    root.children(2).children(0).add(new Node(12))
    root
  }

  def bfs(nodes: List[Node]): List[Node] = {
    if (nodes.nonEmpty) {
      nodes ++ bfs(nodes.flatMap(_.children))
    } else {
      List.empty
    }
  }

  "BFS" should "work" in {
    println(bfs(List(tree())))
  }
}

输出:

List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

愚蠢的方式:

template<typename T>
struct Node { Node* left; Node* right; T value; };

template<typename T, typename P>
bool searchNodeDepth(Node<T>* node, Node<T>** result, int depth, P pred) {
    if (!node) return false;
    if (!depth) {
        if (pred(node->value)) {
            *result = node;
        }
        return true;
    }
    --depth;
    searchNodeDepth(node->left, result, depth, pred);
    if (!*result)
        searchNodeDepth(node->right, result, depth, pred);
    return true;
}

template<typename T, typename P>
Node<T>* searchNode(Node<T>* node, P pred) {
    Node<T>* result = NULL;
    int depth = 0;
    while (searchNodeDepth(node, &result, depth, pred) && !result)
        ++depth;
    return result;
}

int main()
{
    // a c   f
    //  b   e
    //    d
    Node<char*>
        a = { NULL, NULL, "A" },
        c = { NULL, NULL, "C" },
        b = { &a, &c, "B" },
        f = { NULL, NULL, "F" },
        e = { NULL, &f, "E" },
        d = { &b, &e, "D" };

    Node<char*>* found = searchNode(&d, [](char* value) -> bool {
        printf("%s\n", value);
        return !strcmp((char*)value, "F");
    });

    printf("found: %s\n", found->value);

    return 0;
}