Morris Inorder Traversal of a Binary Tree

Problem Statement:

• Example:

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✅ Solution: Morris Inorder Traversal (O(1) Space)

/**
 * 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 {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        TreeNode* current = root;
        vector<int> inorder;

        while (current != NULL) {
            if (current->left == NULL) {
                // No left child: visit current node and move right
                inorder.push_back(current->val);
                current = current->right;
            } else {
                // Find the inorder predecessor of current
                TreeNode* predecessor = current->left;

                while (predecessor->right && predecessor->right != current) {
                    predecessor = predecessor->right;
                }

                if (predecessor->right == NULL) {
                    // Make a thread (temporary link) back to current
                    predecessor->right = current;
                    current = current->left;
                } else {
                    // Thread already exists → remove it and visit current
                    predecessor->right = NULL;
                    inorder.push_back(current->val);
                    current = current->right;
                }
            }
        }
        return inorder;
    }
};

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📝 How It Works

• This is Morris Traversal for inorder, which avoids recursion and stack.
• It leverages temporary right-threaded pointers to backtrack to the root after left subtree.
• For each node:
  • If no left child → visit and move right.
  • If left child exists → find predecessor (rightmost node in left subtree):
    • If thread doesn't exist → create it and move left.
    • If thread exists → remove it, visit current, and move right.

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🧩 Key Idea

• Threads are temporary and restore the tree after traversal.
• Inorder means we visit nodes after exploring their left subtree.

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⏱️ Time & Space Complexity

Metric	Complexity
Time	O(N) – Each node visited at most twice
Space	O(1) – No recursion or extra stack used

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⚠️ Edge Cases

• Empty tree → returns empty vector
• Single node → directly visited
• Skewed trees → still works in constant space

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💡 Other Approaches

Approach	Time	Space	Notes
✅ Morris Traversal	O(N)	O(1)	Most space efficient
Recursive DFS	O(N)	O(H)	Simple but uses stack
Iterative with Stack	O(N)	O(H)	Common and easier to understand

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🔁 Related Problems

• Leetcode 94: Binary Tree Inorder Traversal
• Leetcode 144: Preorder Traversal
• Leetcode 145: Postorder Traversal
• GFG: Morris Inorder Traversal

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