题目

题解

二叉树有点忘记了 专门去复习了 一遍 其实就是 左中右

一看用递归或者栈吧

栈比较简单一些 我们直接不断向左移动 碰到空就向栈顶的右走即可

代码

 1 #include<iostream>
 2 #include<vector>
 3 #include<stack>
 4 using namespace std;
 5 
 6  struct TreeNode {
 7    int val;
 8    TreeNode *left;
 9    TreeNode *right;
10    TreeNode() : val(0), left(nullptr), right(nullptr) {}
11    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
12    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
13  };
14 
15 class Solution {
16 public:
17     vector<int> inorderTraversal(TreeNode* root) {
18         vector<int> result;
19         stack<TreeNode*> ergodic;
20         while (root != nullptr || !ergodic.empty()) {
21             if (root != nullptr) {
22                 ergodic.push(root);
23                 root = root->left;
24             }
25             else
26             {
27                 result.push_back(ergodic.top()->val);
28                 root = ergodic.top()->right;
29                 ergodic.pop();
30             }
31         }
32         return result;
33     }
34 };
35 
36 
37 
38 int main() {
39     Solution sol;
40     //懒得写遍历了
41     TreeNode* root7 = new TreeNode(7);
42     TreeNode* root6 = new TreeNode(6);
43     TreeNode* root5 = new TreeNode(5);
44     TreeNode* root4 = new TreeNode(4);
45     TreeNode* root3 = new TreeNode(3, root6, root7);
46     TreeNode* root2 = new TreeNode(2, root4, root5);
47     TreeNode*  root1=new TreeNode(1, root2, root3);
48     vector<int> result=sol.inorderTraversal(root1);
49     for (int i = 0; i < result.size(); i++) {
50         cout << result[i] << endl;
51     }
52 }

View Code

 

标签：right,TreeNode,中序,nullptr,力扣,二叉树,new,root,left
来源： https://www.cnblogs.com/zx469321142/p/16491458.html