mycode  81.40%

# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution(object):
def kthSmallest(self, root, k):
"""
:type root: TreeNode
:type k: int
:rtype: int
"""
self.res = []
def inorder(root):
if not root:
return
inorder(root.left)
self.res.append(root.val)
inorder(root.right)
inorder(root)
return self.res[k-1]

参考

思路:

我的方法中先将整个tree都遍历了一遍,其实是不必要的,那么该如何恰好在找到第k个数的时候及时退出呢-》

def kthSmallest(root, k):
"""
:type root: TreeNode
:type k: int
:rtype: int
"""
s = []
p = root
cnt = 0
print('p',p)
while s or p:
if p:
s.append(p)
print('if',p.val,s)
p = p.left
else:
p = s[-1].right
cnt += 1 #上面之所以能取出p是因为已经没有左子树了,所以最后左子树的叶子就是目前数里面最小的数,计数+1
print('else',s)
if cnt == k:
return s[-1].val
s.pop()

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