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所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正,讨论交流。
'''
Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?
'''
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def __init__(self):
self.answer = []
def inorderTraversal(self, root: TreeNode) -> List[int]:
# Approach one 递归
# if not root : return []
# return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)
# Approach two
# if not root : return []
# self.inorderTraversal(root.left)
# self.answer.append(root.val)
# self.inorderTraversal(root.right)
# return self.answer
# Approach four 不使用类变量的情况
# def helper(root, res):
# if root:
# helper(root.left, res)
# res.append(root.val)
# helper(root.right, res)
# res = []
# helper(root, res)
# return res
# Approach three 迭代效率稍低
if not root : return []
stack, res = [] , []
while True:
while root:
stack.append(root)
root = root.left
if not stack : return res
node = stack.pop()
res.append(node.val)
root = node.right
所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正,讨论交流。