基于堆栈修改的预置树遍历

Question

我有一个修改的预置树遍历的（nested set model）的递归实现，我想实现而不使用递归。

from collections import deque 

def mptt_recurse(tree, node, preorder=None): 

    if node not in tree: return 
    if preorder is None: preorder = deque() 

    for child in tree[node]: 
     preorder.append(child) 
     mptt_recurse(tree, child, preorder) 
     preorder.append(child) 

    return preorder 

递归执行按预期工作：

>>> tree = { 
    "root": ["food"], 
    "food": ["meat", "veg"], 
    "meat": ["pork", "lamb"], 
    "pork": ["bacon", "sausage"], 
    "lamb": ["cutlet"], 
    "soup": ["leak", "tomato"] 
} 
>>> mptt_recuser(tree, "root") 
deque(['food', 'meat', 'pork', 'bacon', 'bacon', 'sausage', 'sausage', 'pork', 'lamb', 'cutlet', 'cutlet', 'lamb', 'meat', 'veg', 'veg', 'food']) 

每个节点通过在deque位置代表左，右值出现了两次。

def mptt_stack(tree, node): 

    if node not in tree: return 
    stack = deque(tree[node]) 
    preorder = deque() 

    while stack: 
     node = stack.pop() 
     preorder.append(node) 
     if node not in tree: 
      continue 
     for child in reversed(tree[node]): 
      stack.append(child) 

    return preorder 

随着我的堆叠基于实现我只能够弄清楚如何获得序（而不是修改序同时与左，右的标签为每个节点）。

>>> mptt_stack(tree, "root") 
deque(['food', 'meat', 'pork', 'bacon', 'sausage', 'lamb', 'cutlet', 'veg']) 

非递归实现的任何指针都会很好。

Answer 1

这将得到与mptt_recurse相同的结果。它使信息在堆栈上的附加件，这表明无论是进入或离开节点：

def mptt_stack(tree, node): 
    if node not in tree: return 
    preorder = deque() 

    stack = [] 
    for child in reversed(tree[node]): 
     stack.append([child, True]) 

    while stack: 
     (node, first) = stack.pop() 
     preorder.append(node) 
     if first: 
      stack.append([node, False]) 
      if node in tree: 
       for child in reversed(tree[node]): 
        stack.append([child, True]) 

    return preorder 

如果你想在结果初始节点，您可以替换与初始化stack循环：

stack = [[node, True]]