计算二进制搜索树中的重复节点

Question

I am trying to write a function to count the repetitive node in a Binary Search Tree (assume that this tree accept repetition). Here is a line of code that I have

def count(treeroot,item,y):
if treeroot==None:
    return 0
elif treeroot.getData()==item:
    return 1
else:
    return count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)

where y is the starting number of the search (such as search for how many 10 in the tree, we do count(treeroot,10,0)

However, I tried to put 3 number 10 in and I only receive back the count of 1.

Can anyone tell me what is wrong with my code and how to fix it

Answer 1

You stop recursing down the tree as soon as you found the first item. You need to keep recursing:

def count(treeroot,item,y):
    if treeroot==None:
        return 0
    elif treeroot.getData()==item:
        return 1 + count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)
    else:
        return count(treeroot.getLeft(),item,y)+count(treeroot.getRight(),item,y)

I hope you see the problem with your algorithm though: You will visit each and every node of the tree, even if you know that there will be no 10 s to the left of 9 s or to the right of 11 s.