二叉树中序遍历, 迭代法 Python 3 实现：
源代码已上传 Github，持续更新。
"""
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?
"""
"""
二叉树
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:
# 找到最底层的左子节点, 过程中将经过的节点放入栈中
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)
源代码已上传至 Github，持续更新中。
