中序遍历

递归法

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/**
 * 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 {//递归
public:
    void Inorder(TreeNode *root, vector<int> &res){
        if(!root){
            return;
        }
        Inorder(root->left,res);
        res.push_back(root->val);
        Inorder(root->right,res);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        Inorder(root,res);
        return res;
    }


};

迭代法

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/**
 * 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 {//迭代
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> st;
        while(root!=nullptr || !st.empty()){
            while(root!=nullptr){
                st.push(root);
                root=root->left;
            }//这一步算到最后会得到root=null
            root = st.top();//这个即原根节点的最左下的节点
            res.push_back(root->val);
            st.pop();
            root=root->right;
        
        }
        return res;
    }


};

Morris/莫里斯中序遍历

c1b589b5fc7facd1a847c9f5bab407765222ee2d9e1a887a9e5d61cc9e94dfc6-3

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/**
 * 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 {//Morris/莫里斯中序遍历
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        TreeNode *pre;
        while(root!=nullptr){
            if(root->left!=nullptr){
                pre = root->left;
                while(pre->right!=nullptr){
                    pre= pre->right;
                }
                pre->right = root;
                TreeNode *temp = root;
                root=root->left;
                temp->left=nullptr;
            }
            else{
                res.push_back(root->val);
                root = root->right;
            }
            

        }
        
        return res;
    }


};