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# Binary Tree Right Side View

> 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.

## Thoughts

输出二叉树最右侧的一系列元素。BFS，每层把最后一个元素输出。或者先右后左的DFS配合map记录每层最先遇到的元素。

## Code

```cpp
/*
 * @lc app=leetcode id=199 lang=cpp
 *
 * [199] Binary Tree Right Side View
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
         unordered_map<int, int> m;
         int d = -1;
         stack<TreeNode*> s;
         TreeNode *pre = nullptr; 
         const auto push_right = [&](TreeNode *cur) {
             while (cur != nullptr) {
                 ++d;
                 if (!m.count(d)) m[d] = cur->val;
                 s.push(cur);
                 cur = cur->right;
             }
         };
         push_right(root);
         while (!s.empty()) {
             auto t = s.top();
             if (t->left != nullptr && t->left != pre) {
                 push_right(t->left);
                 continue;
             }
             pre = t;
             --d;
             s.pop();
         }
         d = 0;
         vector<int> r;
         while (m.count(d)) {
             r.push_back(m[d++]);
         }
         return r;
    }
};
// @lc code=end


```

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new LinkedList<>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int length = queue.size();
            for (int i = 0; i < length; i++) {
                TreeNode node = queue.poll();
                if (i == length - 1) {
                    res.add(node.val);
                }
                if (node.left != null) queue.add(node.left);
                if (node.right != null) queue.add(node.right);
            }
        }

        return res;
    }
}
```

## Analysis

做题耗时: 6min

时间复杂度O(n)


---

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