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空間是O(1)的Morris Inorder traversal解法。

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 // O(1) space version, Morris Inorder traversal
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        TreeNode* cur = root;
        TreeNode* prev = NULL;
        vector<int> res;
        while(cur) {
            if(cur->left == nullptr) {
                res.push_back(cur -> val);
                cur = cur -> right;
            }
            else {
                prev = cur -> left;
                while(prev -> right != nullptr && prev -> right != cur) {
                    prev = prev -> right;
                }
                if (prev -> right == nullptr) {
                    prev -> right = cur;
                    cur = cur -> left;
                }
                if (prev -> right == cur) {
                    prev -> right = nullptr;
                    res.push_back(cur -> val);
                    cur = cur -> right;
                }
            }
        }
        return res;
    }
};