[LeetCode February Challange] Day 6 - 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.

Example:

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

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

---

Solution

Time complexity : O(n)
Space complexity : O(h)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        ans = {};
        dfs(root, 1);
        return ans;
    }
private:
    vector<int> ans;
    void dfs(TreeNode* root, int level) {
        if (root == nullptr) return;
        if (ans.size() < level)
            ans.push_back(root->val);
        dfs(root->right, level+1);
        dfs(root->left, level+1);
    }
};

用 DFS 且中右左的方式，可以保證在該 level 第一個拜訪的元素是該層最右邊的。