如何在BST中找到第k个最小的节点？ （重访）

Question

I have asked a similar question yesterday but I reached a different solution from the one posted in the original question sO I am reposting with new code. I am not keeping track of number of right and left children of each node. The code works fine for some cases, but for the case of of finding 6th smalest element, it fails. The problem is that I somehow need to carry the number of children down the tree. For example, for node 5, I need to cary over rank of node 4 and I am not able to do that.

This is not a homework, I am trying to prepare for interview and this is one of the classical questions and I can't solve it.

class Node:
    """docstring for Node"""
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
        self.numLeftChildren = 0
        self.numRightChildren = 0


class BSTree:
    def __init__(self):
        # initializes the root member
        self.root = None

    def addNode(self, data):
        # creates a new node and returns it
        return Node(data)

    def insert(self, root, data):
        # inserts a new data
        if root == None:
            # it there isn't any data
            # adds it and returns
            return self.addNode(data)
        else:
            # enters into the tree
            if data <= root.data:
                root.numLeftChildren += 1
                # if the data is less than the stored one
                # goes into the left-sub-tree
                root.left = self.insert(root.left, data)
            else:
                # processes the right-sub-tree
                root.numRightChildren += 1
                root.right = self.insert(root.right, data)
            return root

    def getRankOfNumber(self, root, x, rank):
        if root == None:
            return 0
        if rank == x:
            return root.data
        else:
            if x > rank:
                return self.getRankOfNumber(root.right, x, rank+1+root.right.numLeftChildren)
            if x <= rank:
                return self.getRankOfNumber(root.left, x, root.left.numLeftChildren+1)


# main
btree = BSTree()
root = btree.addNode(13)
btree.insert(root, 3)
btree.insert(root, 14)
btree.insert(root, 1)
btree.insert(root, 4)
btree.insert(root, 18)
btree.insert(root, 2)
btree.insert(root, 12)
btree.insert(root, 10)
btree.insert(root, 5)
btree.insert(root, 11)
btree.insert(root, 8)
btree.insert(root, 7)
btree.insert(root, 9)
btree.insert(root, 6)

print btree.getRankOfNumber(root, 8, rank=root.numLeftChildren+1)

Answer 1

You have the rank of a node. You need to find the rank of its left or right child. Well, how many nodes are between the node and its child?

     a
    / \
   /   \
  b     c
 / \   / \
W   X Y   Z

Here's an example BST. Lowercase letters are nodes; uppercase are subtrees. The number of nodes between a and b is the number of nodes in X . The number of nodes between a and c is the number of nodes in Y . Thus, you can compute the rank of b or c from the rank of a and the size of X or Y .

rank(b) == rank(a) - size(X) - 1
rank(c) == rank(a) + size(Y) + 1

You had the c formula, but the wrong b formula.