[LeetCode] Binary Tree Preorder Traversal

This is a fundamental and yet classic problem. I share my three solutions here:

1. Iterative solution using stack --- O(n) time and O(n) space;
2. Recursive solution --- O(n) time and O(n) space (considering the spaces of call stack);
3. Morris traversal --- O(n) time and O(1) space!!!

Iterative solution using stack:

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

Recursive soltuion:

void preorder(TreeNode* root, vector<int>& nodes) {
    if (!root) return;
    nodes.push_back(root -> val);
    preorder(root -> left, nodes);
    preorder(root -> right, nodes);
} 
vector<int> preorderTraversal(TreeNode* root) {
    vector<int> nodes;
    preorder(root, nodes);
    return nodes;
} 

Morris traversal:

vector<int> preorderTraversal(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)) {
                nodes.push_back(curNode -> val);
                predecessor -> right = curNode;
                curNode = curNode -> left;
            }
            else {
                predecessor -> right = NULL;
                curNode = curNode -> right;
            }
        }
        else {
            nodes.push_back(curNode -> val);
            curNode = curNode -> right;
        }
    }
    return nodes;
}