94. Binary Tree Inorder Traversal

my really bad solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode*> fathers;
        vector<int> res;
        if (root == NULL) return res;
        fathers.push(root);
        while(!fathers.empty()) {
            TreeNode* tmp = fathers.top();
            TreeNode* tmp2 = NULL;
            while (tmp->left) {
                fathers.push(tmp->left);
                tmp2 = tmp;
                tmp = tmp->left;
                tmp2->left = NULL;    //pay attention you need to delete the left node!!
            }
            res.push_back(tmp->val);
            fathers.pop();
            if (tmp->right)
                fathers.push(tmp->right);
        }
        return res;
    }
};

more beautiful one Iterative solution using stack:

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

Morris traversal:

vector<int> inorderTraversal(TreeNode* root) {
    TreeNode* curNode = root;
    vector<int> nodes;
    while (curNode) {
        if (curNode -> left) {
            TreeNode* predecessor = curNode -> left;
            while (predecessor -> right && predecessor -> right != curNode)
                predecessor = predecessor -> right;
            if (!(predecessor -> right)) {
                predecessor -> right = curNode;
                curNode = curNode -> left;
            }
            else {
                predecessor -> right = NULL;
                nodes.push_back(curNode -> val);
                curNode = curNode -> right;
            }
        }
        else {
            nodes.push_back(curNode -> val);
            curNode = curNode -> right;
        }
    }
    return nodes;
}