二叉树层序遍历, Python 3 实现:
源代码已上传 Github,持续更新。
"""
102. Binary Tree Level Order Traversal
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
"""
"""
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 levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
if root == None:
return []
else:
queue = []
result = []
current_node = root
result.append([current_node.val])
queue.append(current_node)
while queue:
sub_queue = []
sub = []
while queue:
current_node = queue[0]
del queue[0]
if current_node.left or current_node.right:
if current_node.left:
sub.append(current_node.left.val)
sub_queue.append(current_node.left)
if current_node.right:
sub.append(current_node.right.val)
sub_queue.append(current_node.right)
queue = sub_queue
if queue:
result.append(sub)
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.levelOrder(root)
源代码已上传至 Github,持续更新中。