软件介绍: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1class TreeNode:def __i...
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)