题目描述
给定一个二叉树的 根节点 root,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
示例 1:
输入: [1,2,3,null,5,null,4] 输出: [1,3,4]
示例 2:
输入: [1,null,3] 输出: [1,3]
个人C++解答
GYL
哈哈,按照自己思路做出来的,果然只要不是递归就好写多了,主要思路就是层次遍历,用数组存每一层,然后保存最后节点的值到ans数组中;
/** * 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) { if (root == nullptr) return {}; vector<TreeNode*> q; vector<int> ans; q.push_back(root); while (!q.empty()) { vector<TreeNode*> vec; for (int i = 0; i <q.size(); i++) { TreeNode* node = q[i]; if (node->left) vec.push_back(node->left); if (node->right) vec.push_back(node->right); } ans.push_back(q[q.size()-1]->val); q = move(vec); } return ans; } };