Question

我正在尝试使用DFS而不是BFS编写返回树中最深叶子的深度的代码，每个节点具有任意数量的子节点。它看起来很接近，但下面的代码仍有一些我无法弄清楚的错误（即返回的深度不正确）。有什么帮助吗？

测试树只是：[[1,2,3]，[4,5]，[6]，[7]，[8]，[]，[]，[]，[]] < / p>

def max_depth_dfs(tree): # DOESN'T WORK

    max_depth, curr_depth, Q = 0,0, [0]
    visited = set()

    while Q != []:
        n = Q[0]
        more = [v for v in tree[n] if v not in visited]
        if not more:
            visited.add(n)
            curr_depth -= 1
            Q = Q[1:]
        else:
            curr_depth += 1

        max_depth = max(max_depth, curr_depth)
        Q = more + Q

    return max_depth

Answer 1

我使用try .. catch来区分树枝和树叶。更新不再有例外：）

from collections import Iterable
tree = [[1,2,3],[4,5, [1, 6]],[6],[7],[8],[],[],[],[]]

def max_depth(tree, level=0):
  if isinstance(tree, Iterable):
    return max([ max_depth(item, level+1) for item in tree])
  else: # leaf
    return level

print max_depth(tree)

Answer 2

我发现了这个错误！

if not more:
    visited.add(n)
    curr_depth -= 1
    Q = Q[1:]

当您访问节点4时，curr_depth等于2.节点4没有子节点，因此您减少curr_depth并且curr_depth现在等于1。但是，您将访问的下一个节点是节点5，节点5的深度是2而不是1.因此，curr_depth不会记录树中节点的正确深度。

以下解决方案可能会有所帮助。

def max_depth_dfs(tree):

    max_depth, curr_depth, Q = 0, 0, [0]
    visited = set()

    while Q != []:
        n = Q[0]

        max_depth = max(max_depth, curr_depth)

        if n in visited:
            curr_depth -= 1
            Q = Q[1:]
            continue

        #print n, curr_depth     #show the node and its depth in the tree

        visited.add(n)
        more = [v for v in tree[n]]
        if not more:
            Q = Q[1:]
        else:
            curr_depth += 1
            Q = more + Q

    return max_depth

Answer 3

这是非复黑版本：

from collections import Iterable

def max_depth_no_recur(tree):
  max_depth, node =  0, iter(tree)
  stack = [node]
  while stack:
    try:
      n = node.next()
    except StopIteration:
      if len(stack) > max_depth:
        max_depth = len(stack)
      node = stack.pop()
      continue

    if isinstance(n, Iterable):
        stack.append(node)
        node = iter(n)
  return max_depth

Answer 4

考虑到我从Alex和Adonis获得的所有良好反馈并完善代码后，我目前拥有当前版本：

def max_depth_dfs(tree): # correct

max_depth, curr_depth, Q = 0, 0, [0]
visited = set()

while Q != []:

    n = Q[0]

    if n in visited:
        Q = Q[1:]
        curr_depth -= 1
        visited.remove(n) # won't go back, save memory 
        print 'backtrack from', n        
        continue

    # proper place to print depth in sync with node id
    print 'visiting', n, 'children=', tree[n], 'curr_depth=', curr_depth, 'Q=', Q,
    print visited # only current path, instead of visited part of tree

    if tree[n]:
        visited.add(n) # if leaf, won't ever try to revisit
        Q = tree[n] + Q
        curr_depth += 1
        max_depth = max(max_depth, curr_depth) # no need to check if depth decreases
    else:
        Q = Q[1:] # leaf: won't revisit, will go to peer, if any, so don't change depth
        print 'no children for', n

return max_depth

使用DFS的树的深度

4 个答案: