Question

首先，我知道还有很多其他线程处理这个问题。我想知道的是，当我在CodeEval上提交时，我的解决方案似乎没有达到100％的排名。它只排在70％。当我在我的本地环境中测试它时，它适用于我所投入的所有测试情况。

我知道我的解决方案并不是最有效的。我试图做的是自己想出一个解决方案，我会理解的。我最终会将它与其他人进行比较。但在此之前，有人可以解释一下我在这里做错了吗？

class binary_tree(object):
def __init__(self, value=None, left=None, right=None):
    self.value=value
    self.left=left
    self.right=right

def insert(self, num):
    #print "...now:", num
    if self.value==num:
        return
    else:
        #print "descending from", self.value
        if num<self.value:
            if self.left:
                #print "descending to left"
                self.left.insert(num)
            else:
                #print "! found empty left"
                self.left=binary_tree(num)
        else:
            if self.right:
                #print "descending to right"
                self.right.insert(num)
            else:
                #print "! found empty right"
                self.right=binary_tree(num)

def tree_from_list(self, value, numbers):
    self.value=value
    for num in numbers:
        self.insert(num)

def __repr__(self,depth=0):
    rtn=""
    rtn+="\t"*depth+" "+str(self.value)+"\n"
    depth+=1
    if self.left:
        rtn+="\t"*depth+" left: "
        rtn+=self.left.__repr__(depth)
    if self.right:
        rtn+="\t"*depth+" right: "
        rtn+=self.right.__repr__(depth)
    return rtn


def find_parent_depth(self, num, depth=0):
    if self.left and self.left.value==num:
        #returns a list of two values, the first one being
        #the depth of the parent, and the second the value
        #itself. depth starts at 0, and increases as we descend
        return [depth, self.value]
    elif self.right and self.right.value==num:
        return [depth, self.value]
    else:
        depth+=1
        #checks for which path to descend on
        if num<self.value:
            if self.left:
                return self.left.find_parent_depth(num, depth)
            else:
                return self.value
        else:
            if self.right:
                return self.right.find_parent_depth(num, depth)
            else:
                return self.value

#lca = lowest common ancestor
def lca(self, v1, v2):  
    parent1=self.find_parent_depth(v1)
    parent2=self.find_parent_depth(v2)
    #checks for which parent has lower depth
    if parent1[0]<parent2[0]:
        #searches for ancestors till it reaches the same depth level
        while parent1[0]!=parent2[0]: 
            parent2=self.find_parent_depth(parent2[1])
        #checks if the ancestors coincide at that depth
        if parent1[1]==parent2[1]:
            #if they do, returns the parent value
            #THIS IS IT
            return parent1[1]
        #if it doesn't, we need to raise the level of the lowest one 
        #and do it all over again
        else:
            return self.lca(parent1[1], parent2[1])
    else:
        #searches for ancestors till it reaches the same depth level
        while parent2[0]!=parent1[0]: 
            parent1=self.find_parent_depth(parent1[1])
        #checks if the ancestors coincide at that depth
        if parent1[1]==parent2[1]:
            #if they do, returns the parent value
            #THIS IS IT
            return parent1[1]
        #if it doesn't, we need to raise the level of the lowest one 
        #and do it all over again
        else:
            return self.lca(parent2[1], parent1[1])
dis=binary_tree()
dis.tree_from_list(30, [8, 52, 3, 20, 10, 29, 12, 90, 89, 1])
print dis.lca(89, 12)

Answer 1

我有类似的问题。鉴于：

   30
   _|___
  |    |
  8   52
 _|___
 |   |
 3  20
    _|___
    |   |
   10   29

我的挂断与LCA定义有关。

我发现20和29的最低共同祖先是20，而不是8，正如我原先认为的那样，给出了方向上的例子。另一个例子是8和25的LCA是8.

调整代码后，我的提交通过了100％。

希望这会有所帮助。干杯！

Python：二叉树中最低的共同祖先，对CodeEval的挑战

1 个答案: