[leetcode: Python]107.Binary Tree Level Order Traversal 2

最新推荐文章于 2022-07-27 15:36:05 发布

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最新推荐文章于 2022-07-27 15:36:05 发布

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分类专栏：算法文章标签： python leetcode

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本文介绍了一种从叶子节点到根节点的二叉树层序遍历算法，并提供了两种实现方法，一种使用递归预序遍历，另一种采用栈进行深度优先搜索。

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题目：
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

这个可以直接把另一道二叉树层序遍历的结果倒一转

方法一：性能62ms

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def preOrder(self, root, level, res):
        if root:
            if len(res)<level+1:
                res.append([])
            res[level].append(root.val)
            self.preOrder(root.left, level+1, res)
            self.preOrder(root.right, level+1, res)


    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        res = []
        self.preOrder(root, 0, res)
        #res.reverse()
        res.reverse()
        return res

方法二：性能45ms

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):

    # # BFS
    # def levelOrderBottom(self, root):
    #     """
    #     :type root: TreeNode
    #     :rtype: List[List[int]]
    #     """
    #     if not root:
    #         return []
    #     res = []
    #     queue = collections.deque([(root, 0)])
    #     while queue:
    #         node, level = queue.popleft()
    #         if node:
    #             if len(res)<level+1:
    #                 res.insert(0,[])
    #             res[-(level+1)].append(node.val)
    #             queue.append((node.left,level+1))
    #             queue.append((node.right,level+1))
    #     return res

    # DFS + stack
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []

        res = []
        stack = [(root,0)]
        while stack:
            node,level = stack.pop()
            if node:
                if len(res)<level+1:
                    res.insert(0,[])
                res[-(level+1)].append(node.val)
                stack.append((node.right,level+1))
                stack.append((node.left,level+1))
        return res

第二种方法再好好看看！