Leetcode 199. Binary Tree Right Side View

Description: 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.

Link: 199. Binary Tree Right Side View

Examples:

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: []

 

思路: 逐层遍历，取每层最后一个元素。所以逐层遍历既可以用DFS, 也可以用BFS。

class Solution(object):
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root: return []
        res = []
        self.dfs(root, 0, res)
        return [l[-1] for l in res]
    
    def dfs(self, root, level, res):
        if not root: return
        if len(res) == level:
            res.append([])
        res[level].append(root.val)
        if root.left:
            self.dfs(root.left, level+1, res)
        if root.right:
            self.dfs(root.right, level+1, res)

BFS:

class Solution(object):
    def rightSideView(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root: return []
        res = []
        que = collections.deque()
        que.append(root)
        while que:
            size = len(que)
            level = []
            for _ in range(size):
                node = que.popleft()
                if not node:
                    continue
                level.append(node.val)
                que.append(node.left)
                que.append(node.right)
            res.append(level)
        # print(res[:-1])   
        return [l[-1] for l in res[:-1]]

日期: 2021-03-26 阴天