Python实现查找二叉搜索树第k大的节点功能示例

本文实例讲述了Python实现查找二叉搜索树第k大的节点功能。分享给大家供大家参考，具体如下：

题目描述

给定一个二叉搜索树，找出其中第k大的节点

就是一个中序遍历的过程，不需要额外的数组，便利到节点之后，k减一就行。

代码1

class TreeNode:
  def __init__(self, x):
    self.val = x
    self.left = None
    self.right = None
class Solution:
  def __init__(self):
    self.k = 0
  def recursionKthNode(self, Root):
    result = None
    if result == None and Root.left:
      result = self.recursionKthNode(Root.left)
    if result == None:
      if self.k == 1:
        return Root
      self.k -= 1
    if result == None and Root.right:
      result = self.recursionKthNode(Root.right)
    return result
  def KthNode(self, Root, k):
    if Root == None:
      return None
    self.k = k
    return self.recursionKthNode(Root)
Root = TreeNode(5)
Root.left = TreeNode(3)
Root.left.left = TreeNode(2)
Root.left.right = TreeNode(4)
Root.right = TreeNode(7)
Root.right.left = TreeNode(6)
Root.right.right = TreeNode(8)
print(Solution().KthNode(Root,3).val)

output : 4

代码2

class TreeNode:
  def __init__(self, x):
    self.val = x
    self.left = None
    self.right = None
class Solution:
  def __init__(self):
    self.k = 0
  def InOrder(self, Root):
    ans = None
    if Root:
      if ans == None and Root.left:
        ans = self.InOrder(Root.left)  #往左遍历
      if ans == None and self.k == 1:
        ans = Root           #遍历到目标节点
      if ans == None and self.k != 1:   #没有遍历到目标节点，k--
        self.k -= 1
      if ans == None and Root.right:   #往右遍历
        ans = self.InOrder(Root.right)
    return ans
  def KthNode(self, Root, k):
    if Root == None or k <= 0:
      return None
    self.k = k
    return self.InOrder(Root)

希望本文所述对大家Python程序设计有所帮助。