[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:

 1 class Solution {
 2 public:
 3     vector<int> inorderTraversal(TreeNode* root) {
 4         vector<int> nodes;
 5         TreeNode* curNode = root;
 6         stack<TreeNode*> toVisit;
 7         while (curNode || !toVisit.empty()) {
 8             if (curNode) {
 9                 toVisit.push(curNode);
10                 curNode = curNode -> left;
11             }
12             else {
13                 curNode = toVisit.top();
14                 toVisit.pop();
15                 nodes.push_back(curNode -> val);
16                 curNode = curNode -> right;
17             }
18         }
19         return nodes;
20     }
21 }; 

Recursive solution:

 1 class Solution {
 2 public:
 3     void inorder(TreeNode* node, vector<int>& nodes) {
 4         if (!node) return;
 5         inorder(node -> left, nodes);
 6         nodes.push_back(node -> val);
 7         inorder(node -> right, nodes);
 8     }
 9     vector<int> inorderTraversal(TreeNode* root) {
10         vector<int> nodes;
11         inorder(root, nodes);
12         return nodes;
13     }
14 }; 

Morris traversal:

 1 class Solution {
 2 public:
 3     vector<int> inorderTraversal(TreeNode* root) {
 4         vector<int> nodes;
 5         TreeNode* curNode = root;
 6         while (curNode) {
 7             if (curNode -> left) {
 8                 TreeNode* predecessor = curNode -> left;
 9                 while (predecessor -> right && predecessor -> right != curNode)
10                     predecessor = predecessor -> right;
11                 if (predecessor -> right == NULL) {
12                     predecessor -> right = curNode;
13                     curNode = curNode -> left;
14                 }
15                 else {
16                     predecessor -> right = NULL;
17                     nodes.push_back(curNode -> val);
18                     curNode = curNode -> right;
19                 }
20             }
21             else {
22                 nodes.push_back(curNode -> val);
23                 curNode = curNode -> right;
24             }
25         }
26         return nodes;
27     }
28 };

 

posted @ 2015-06-02 23:22  jianchao-li  阅读(254)  评论(0)  编辑  收藏  举报