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  • Find Leaves of Binary Tree

    Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

    Example:
    Given binary tree 

              1
             / 
            2   3
           /      
          4   5    
    

    Returns [4, 5, 3], [2], [1].

    Explanation:

    1. Removing the leaves [4, 5, 3] would result in this tree:

              1
             / 
            2          
    

    2. Now removing the leaf [2] would result in this tree:

              1          
    

    3. Now removing the leaf [1] would result in the empty tree:

              []         
    

    Returns [4, 5, 3], [2], [1].

    很有意思的一道题,主要是一步一步去除叶子结点,然后返回结果。其实这题考的是二叉树结点的高度。高度最低的先输出,同等高度的一块输出。略微修改下递归球二叉树高度的方法就可以,同时可以根据高度,确定加结果的地方。空间复杂度O(logn),时间复杂度位O(n),每个节点都需要处理一次。代码如下:

    class Solution(object):
        def findLeaves(self, root):
            """
            :type root: TreeNode
            :rtype: List[List[int]]
            """
            if not root:
                return []
            res = []
            self.dfs(root, res)
            return res
        def dfs(self, root, res):
            if not root:
                return -1
            left = self.dfs(root.left, res)
            right = self.dfs(root.right, res)
            level = max(left, right) + 1
            if len(res) == level:
                res.append([root.val])
            else:
                res[level].append(root.val)
            root.left = root.right = None
            return level
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  • 原文地址:https://www.cnblogs.com/sherylwang/p/5667583.html
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