一、Leetcode-199：二叉树的右视图

1.1 问题描述

1.2 问题分析

这道题还是基于按层遍历的基础上，将每一层的最后一个元素push_back到vector中就可以。

1.3 问题解决

/*** 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) {vector<vector<int>> level;queue<TreeNode*> q;if(root) q.push(root);while(!q.empty()){int n = q.size();vector<int> v;for(int i = 0; i < n; i++){TreeNode* node = q.front();q.pop();v.push_back(node->val);if(node->left) q.push(node->left);if(node->right) q.push(node->right);}level.push_back(v);}vector<int> ans;for(int i = 0; i < level.size(); i++)ans.push_back(level[i][level[i].size()-1]);return ans;}
};

二、Leetcode-1028：从先序遍历还原二叉树

2.1 问题描述

2.2 问题分析

这道题题解看懂了，但是代码不知道怎么写，遂先放一放，抄了一下题解。

2.3 问题解决

/*** 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)s {}* };*/
class Solution {
public:TreeNode* recoverFromPreorder(string str) {stack<TreeNode*> path;int pos = 0;while(pos < str.size()){int level = 0;while(str[pos] == '-') { level++; pos++;}int val = 0;while(pos < str.size() && isdigit(str[pos])){val = val * 10 + str[pos] - '0';pos++;}TreeNode* node = new TreeNode(val);if(level == path.size()){if(!path.empty()) path.top()->left = node;}else{while(level != path.size()) path.pop();path.top()->right = node;}path.push(node);}while(path.size() > 1) path.pop();return path.top();}
};