力扣 题目94- 二叉树的中序遍历

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

#include<iostream>
#include<vector>
#include<stack>
using namespace std;

 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> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> ergodic;
        while (root != nullptr || !ergodic.empty()) {
            if (root != nullptr) {
                ergodic.push(root);
                root = root->left;
            }
            else
            {
                result.push_back(ergodic.top()->val);
                root = ergodic.top()->right;
                ergodic.pop();
            }
        }
        return result;
    }
};



int main() {
    Solution sol;
    //懒得写遍历了
    TreeNode* root7 = new TreeNode(7);
    TreeNode* root6 = new TreeNode(6);
    TreeNode* root5 = new TreeNode(5);
    TreeNode* root4 = new TreeNode(4);
    TreeNode* root3 = new TreeNode(3, root6, root7);
    TreeNode* root2 = new TreeNode(2, root4, root5);
    TreeNode*  root1=new TreeNode(1, root2, root3);
    vector<int> result=sol.inorderTraversal(root1);
    for (int i = 0; i < result.size(); i++) {
        cout << result[i] << endl;
    }
}