你可以使用一个堆栈来保存尚未访问的节点:

stack.push(root)
while !stack.isEmpty() do
    node = stack.pop()
    for each node.childNodes do
        stack.push(stack)
    endfor
    // …
endwhile

基于biziclops的ES6实现很棒的答案:

root = { text: "root", children: [{ text: "c1", children: [{ text: "c11" }, { text: "c12" }] }, { text: "c2", children: [{ text: "c21" }, { text: "c22" }] }, ] } console.log("DFS:") DFS(root, node => node.children, node => console.log(node.text)); console.log("BFS:") BFS(root, node => node.children, node => console.log(node.text)); function BFS(root, getChildren, visit) { let nodesToVisit = [root]; while (nodesToVisit.length > 0) { const currentNode = nodesToVisit.shift(); nodesToVisit = [ ...nodesToVisit, ...(getChildren(currentNode) || []), ]; visit(currentNode); } } function DFS(root, getChildren, visit) { let nodesToVisit = [root]; while (nodesToVisit.length > 0) { const currentNode = nodesToVisit.shift(); nodesToVisit = [ ...(getChildren(currentNode) || []), ...nodesToVisit, ]; visit(currentNode); } }

使用堆栈来跟踪节点

Stack<Node> s;

s.prepend(tree.head);

while(!s.empty) {
    Node n = s.poll_front // gets first node

    // do something with q?

    for each child of n: s.prepend(child)

}

DFS:

list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
  currentnode = nodes_to_visit.take_first();
  nodes_to_visit.prepend( currentnode.children );
  //do something
}

BFS:

list nodes_to_visit = {root};
while( nodes_to_visit isn't empty ) {
  currentnode = nodes_to_visit.take_first();
  nodes_to_visit.append( currentnode.children );
  //do something
}

两者的对称相当酷。

更新:如前所述，take_first()删除并返回列表中的第一个元素。

你可以使用一个堆栈来保存尚未访问的节点:

stack.push(root)
while !stack.isEmpty() do
    node = stack.pop()
    for each node.childNodes do
        stack.push(stack)
    endfor
    // …
endwhile

伪代码基于@biziclop的答案:

只使用基本结构:变量、数组、if、while和for 函数getNode(id)和getChildren(id) 假设已知节点数N

注意:我从1开始使用数组索引，而不是0。

广度优先

S = Array(N)
S[1] = 1; // root id
cur = 1;
last = 1
while cur <= last
    id = S[cur]
    node = getNode(id)
    children = getChildren(id)

    n = length(children)
    for i = 1..n
        S[ last+i ] = children[i]
    end
    last = last+n
    cur = cur+1

    visit(node)
end

深度优先

S = Array(N)
S[1] = 1; // root id
cur = 1;
while cur > 0
    id = S[cur]
    node = getNode(id)
    children = getChildren(id)

    n = length(children)
    for i = 1..n
        // assuming children are given left-to-right
        S[ cur+i-1 ] = children[ n-i+1 ] 

        // otherwise
        // S[ cur+i-1 ] = children[i] 
    end
    cur = cur+n-1

    visit(node)
end