二叉树的遍历

最新推荐文章于 2024-10-21 08:57:50 发布

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最新推荐文章于 2024-10-21 08:57:50 发布

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分类专栏： python 文章标签：二叉树算法

本文链接：https://blog.youkuaiyun.com/qq_36711094/article/details/115904791

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前序：根左右的顺序进行遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        return [root.val]+self.preorderTraversal(root.left)+self.preorderTraversal(root.right)

中序：左根右的顺序进行遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        return self.inorderTraversal(root.left)+[root.val]+self.inorderTraversal(root.right)

后序：左右根的顺序进行遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        if root is None:
            return []
        return  self.postorderTraversal(root.left) + self.postorderTraversal(root.right) + [root.val]

层序：每层遍历，设置队列进行每层存储到temp，temp汇总到res，再将最后的res输出

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        res = []
        q = collections.deque()
        q.append(root)
        while q:
            temp = []
            for _ in range(len(q)):
                node = q.popleft()
                # 头插
                temp.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            res.append(temp)
        
        return res