问题：
给定二叉树，求从右边观察这棵树，能看到的每层第一个元素。
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: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100
解法：BSF，DFS
解法一：BFS
queue：每层节点
for每层节点遍历。for结束，将最后一个节点res.push_back
代码参考：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 vector<int> rightSideView(TreeNode* root) { 15 vector<int> res; 16 queue<TreeNode*> q; 17 if(root) q.push(root); 18 while(!q.empty()) { 19 int sz = q.size(); 20 TreeNode* cur; 21 for(int i=0; i<sz; i++) { 22 cur = q.front(); 23 q.pop(); 24 if(cur->left) q.push(cur->left); 25 if(cur->right) q.push(cur->right); 26 } 27 res.push_back(cur->val); 28 } 29 return res; 30 } 31 };
解法二：DFS
状态：当前层：若访问新的层(res.size<level) 将当前节点res.push_back
⚠️ 注意：因为要每次访问新层的第一个节点push到res中，那么就要从右向左遍历。
对于当前节点进行，子节点递归时：
- 先 node->right
- 再 node->left
递归退出条件：node==null
代码参考：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val(0), left(nullptr), right(nullptr) {} 8 * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} 9 * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} 10 * }; 11 */ 12 class Solution { 13 public: 14 void dfs(vector<int>& res, int level, TreeNode* node) { 15 if(!node) return; 16 if(level>res.size()) res.push_back(node->val); 17 //fist right-> then left ← 18 dfs(res, level+1, node->right); 19 dfs(res, level+1, node->left); 20 return; 21 } 22 vector<int> rightSideView(TreeNode* root) { 23 vector<int> res; 24 dfs(res, 1, root); 25 return res; 26 } 27 };