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;
}
};

