[LeetCode] 199. Binary Tree Right Side View_ Medium tag: BFS, Amazon

Given 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:

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

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

这个题目也是经典的BFS, 依次scan tree的每一层, 然后将第一个看到的元素append进入ans里面即可. 这里思路是根据ans的size来判断该层是否已经append过了, 如果append过了那么就继续测试他的children. 最后返回ans.

1. Constraints

1) Tree 可以为empty, 但是因为返回的是array, 所以可以将edge case放到queue里面去判断即可.

2. Ideas

BFS     T: O(n),  S: O(n)   n is number of nodes of the tree

1) ans(init: []), queue( init: [root, 0])

2) while queue:  node, heig = queue.popleft(), 根据len(ans), 和heig的大小来判断该层是否被append过, 如果没有, 那么append进入ans

3) 先判断right children, 因为是right view, 那么同理如果出一个题目, 问的是left view, 那么code基本不变, 只是将判断right children 和left children的顺序换下即可.

4) return ans

3. code

 class Solution:
     def rightSideView(self, root):
         ans, queue = [], collections.deque([(root, 0)])
         while queue:
             node, heig = queue.popleft()
             if node:
                 if heig == len(ans):
                     ans.append(node.val)
                 if node.right:
                     queue.append((node.right, heig + 1))
                 if node.left:
                     queue.append((node.left, heig + 1))
         return ans

similar, if we change the question into Binary Tree Left Side View. then the code is similar, but change the order to append left and right child.

 class Solution:
     def leftSideView(self, root):
         ans, queue = [], collections.deque([(root, 0)])
         while queue:
             node, heig = queue.popleft()
             if node:
                 if heig == len(ans):
                     ans.append(node.val)
                 if node.left:
                     queue.append((node.left, heig + 1))
                 if node.right:
                     queue.append((node.right, heig + 1))
         return ans

4. Test cases

1) empty tree

2)

   1            <---
 /   \
2     3         <---
 \
  5            <---

3)

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---