Python: 二叉树中的最近公共祖先，CodeEval挑战

Question

首先，我知道有很多其他的讨论在解决这个问题。我想知道的是，为什么我提交的解决方案在CodeEval上只得了70%的分数，而不是100%。在我本地测试的时候，它在我给它的所有测试情况下都能正常工作。

我知道我的解决方案不是最有效率的。我尝试自己想出一个我能理解的解决方案。最终我会和其他人的方案进行比较。但在那之前，有人能告诉我我哪里做错了吗？

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

我遇到过类似的问题。

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

我卡住的地方是关于最低公共祖先的定义。

我发现20和29的最低公共祖先是20，而不是我最开始根据说明中的例子认为的8。另一个例子是8和20的最低公共祖先是8。

调整代码后，我的提交通过了100%。

希望这对你有帮助。加油！