二叉樹(shù)中序遍歷, 迭代法 Python 3 實(shí)現(xiàn)：
源代碼已上傳 Github，持續(xù)更新。
"""
94. Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
"""
"""
二叉樹(shù)
0
/ \
1 2
/ \ / \
3 4 5 6
"""
# Definition for a binary tree node.
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution:
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if root == None:
return []
else:
stack = []
result = []
current_node = root
while current_node or stack:
# 找到最底層的左子節(jié)點(diǎn), 過(guò)程中將經(jīng)過(guò)的節(jié)點(diǎn)放入棧中
while current_node:
stack.append(current_node)
current_node = current_node.left
current_node = stack.pop()
result.append(current_node.val)
current_node = current_node.right
return result
if __name__ == '__main__':
root = TreeNode(0)
node1 = TreeNode(1)
node2 = TreeNode(2)
node3 = TreeNode(3)
node4 = TreeNode(4)
node5 = TreeNode(5)
node6 = TreeNode(6)
root.left = node1
root.right = node2
node1.left = node3
node1.right = node4
node2.left = node5
node2.right = node6
solution = Solution()
solution.inorderTraversal(root)
源代碼已上傳至 Github，持續(xù)更新中。
