首先需要确定树的高度，再利用递归的思想，先遍历右子树，再遍历左子树。
在遍历过程中，要检查当前层是否访问过，以及当前层高度是否等于树高
/*** 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:int getHeight(TreeNode* root){ int level = 0;TreeNode* cur;deque<TreeNode*> queue;if (root == NULL) return 0;/* 根节点入队 */queue.push_back(root);/* 遍历 */while (!queue.empty()){int size = queue.size();level++;for (int i = 0; i < size; i++){cur = queue.front();queue.pop_front();if (cur->left != NULL)queue.push_back(cur->left);if (cur->right != NULL)queue.push_back(cur->right);}}return level;}vector<int> rightSideView(TreeNode* root) { vector<int> res;int curHeight = 0;int curMaxHeight = 0;int height = getHeight(root);viewFromRight(root, curHeight, curMaxHeight, height, res);return res;}void viewFromRight(TreeNode* root, int& curHeight, int& curMaxHeight, int height, vector<int>& res){ if (root == NULL) return;curHeight++;//根节点if (curHeight == 1){curMaxHeight = curHeight;res.push_back(root->val);viewFromRight(root->right, curHeight, curMaxHeight, height, res);viewFromRight(root->left, curHeight, curMaxHeight, height, res);}//已完成遍历else if (curMaxHeight == height){return;}//非根节点，且当前层未访问过else if (curHeight > curMaxHeight){curMaxHeight = curHeight;res.push_back(root->val);viewFromRight(root->right, curHeight, curMaxHeight, height, res);viewFromRight(root->left, curHeight, curMaxHeight, height, res);}//非根节点，且当前层访问过else if (curHeight <= curMaxHeight){viewFromRight(root->right, curHeight, curMaxHeight, height, res);viewFromRight(root->left, curHeight, curMaxHeight, height, res);}curHeight--;}
};