[LeetCode] Binary Tree Inorder 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 costs of call stack);
3. Morris traversal --- O(n) time and O(1) space!!!

---

Iterative solution using stack:

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nodes;
        TreeNode* curNode = root;
        stack<TreeNode*> toVisit;
        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;
    }
}; 

---

Recursive solution:

class Solution {
public:
    void inorder(TreeNode* node, vector<int>& nodes) {
        if (!node) return;
        inorder(node -> left, nodes);
        nodes.push_back(node -> val);
        inorder(node -> right, nodes);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nodes;
        inorder(root, nodes);
        return nodes;
    }
}; 

---

Morris traversal:

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nodes;
        TreeNode* curNode = root;
        while (curNode) {
            if (curNode -> left) {
                TreeNode* predecessor = curNode -> left;
                while (predecessor -> right && predecessor -> right != curNode)
                    predecessor = predecessor -> right;
                if (predecessor -> right == NULL) {
                    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;
    }
};