leetcode--94.二叉树的中序遍历

记录
2024-1-21

https://leetcode.cn/problems/binary-tree-inorder-traversal/

二叉树的中序遍历，递归很简单，迭代需要利用栈，当第二次遇到元素的时候再加入到结果中(这句话除我之外可能别人都看不懂，因为这和我之前写过的博客有关->https://www.cnblogs.com/57one/p/11423218.html)

/**
 * 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) {}
 * };
 */
#include <stack>
using namespace std;
class Solution {
public:
    // vector<int> inorderTraversal(TreeNode* root) {
    //     if(root==nullptr) {
    //         return {};
    //     }

    //     vector<int> result, lch, rch;
    //     if(root->left!=nullptr) {
    //         lch = inorderTraversal(root->left);
    //         result = lch;
    //     }

    //     result.push_back(root->val);
    //     if(root->right!=nullptr) {
    //         rch = inorderTraversal(root->right);
    //         result.insert(result.end(), rch.begin(), rch.end());
    //     }
    //     return result;
    // }


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