[leetcode]199. Binary Tree Right Side View二叉树右视图

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       <---

思路

DFS

每当recursion一进入到next level，

就立马加上该level的right side node到result里

对应的，

在recursion的时候，先处理 root.right

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代码

 1 class Solution {
 2     public List<Integer> rightSideView(TreeNode root) {
 3         List<Integer> result = new ArrayList<>();
 4         if(root == null) return result;
 5         dfs(root, result, 0);
 6         return result;     
 7     }
 8     
 9     private void dfs(TreeNode root, List<Integer> result, int level ){
10         // base case
11         if(root == null) return;
12         /*height == result.size() limits the amount of Node add to the result
13         making sure that once go to the next level, add right side node to result immediately
14         */
15         if(level == result.size()){
16             result.add(root.val);
17         }
18         // deal with right side first, making sure right side node to be added first
19         dfs(root.right, result, level+1);
20         dfs(root.left, result, level+1);
21     }
22 }

思路

BFS（iteration）

每次先将right side node 加入到queue里去

保证 当i = 0 的时候，poll出来的第一个item是right side node

代码

 1 public List<Integer> rightSideView(TreeNode root) {
 2         // level order traversal
 3         List<Integer> result = new ArrayList();
 4         Queue<TreeNode> queue = new LinkedList();
 5         // corner case
 6         if (root == null) return result;
 7         
 8         queue.offer(root);
 9         while (!queue.isEmpty()) {
10             int size = queue.size();
11             for (int i = 0; i< size; i++) {
12                 TreeNode cur = queue.poll();
13                 // make sure only add right side node
14                 if (i == 0) result.add(cur.val);
15                 // add right side node first, making sure poll out first
16                 if (cur.right != null) queue.offer(cur.right);
17                 if (cur.left != null) queue.offer(cur.left);
18             }
19         }
20         return result;
21     }