leetcode 199. Binary Tree Right Side View（python）

描述

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Note:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

解析

根据题意，给定二叉树的根 root ，想象你站在它的右边，返回你可以看到从上到下能看到的最右边的节点的值。

因为我们要找到每一层的最右边的节点值，所以我们要按照层进行遍历节点，将所有最右边的值放入结果 result 中皆可，其实这道题考察的就是使用使用 BFS 进行层序遍历。

时间复杂度为 O(N) ，空间复杂度为 O(N) 。

解答

class Solution(object):
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        q = [root]
        result = [root.val]
        while q:
            l =  len(q)
            for i in range(l):
                node = q.pop(0)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            if q:
                result.append(q[-1].val)
        return result

运行结果

Runtime: 24 ms, faster than 70.72% of Python online submissions for Binary Tree Right Side View.
Memory Usage: 13.4 MB, less than 52.08% of Python online submissions for Binary Tree Right Side View.

原题链接

leetcode.com/problems/bi…

您的支持是我最大的动力