Add Solution for 094 Binary Tree Inorder Traversal

cpp/094_Binary_Tree_Inorder_Traversal.cpp

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+// 94. Binary Tree Inorder Traversal
+/**
+ * Given a binary tree, return the inorder traversal of its nodes' values.
+ * 
+ * For example:
+ * Given binary tree [1,null,2,3],
+ *    1
+ *     \
+ *      2
+ *     /
+ *    3
+ * return [1,3,2].
+ * 
+ * Note: Recursive solution is trivial, could you do it iteratively?
+ * 
+ * Tags: Tree, Hash Table, Stack
+ * 
+ * Similar Problems: (M) Validate Binary Search Tree, (M) Binary Tree Preorder Traversal, (H) Binary Tree Postorder Traversal, 
+ * (M) Binary Search Tree Iterator (M) Kth Smallest Element in a BST (H) Closest Binary Search Tree Value II (M) Inorder Successor in BST
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+#include <vector>
+#include <stack>
+
+using namespace std;
+
+/**
+ * Definition for a binary tree node.
+ */
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+    TreeNode(int x, TreeNode* l, TreeNode* r) : val(x), left(l), right(r) {}
+};
+
+// recursive solution
+// time: O(n), space: O(n) - function call stack
+class Solution {
+    void inorder(TreeNode* root, vector<int>& nodes) {
+        if (root == NULL) {
+            return;
+        }
+
+        inorder(root->left, nodes);
+        nodes.push_back(root->val);
+        inorder(root->right, nodes);
+        return;
+    }
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> res;
+        inorder(root, res);
+        return res;
+    }
+};
+
+// iterative solution using stack
+// time: O(n), space: O(n) - stack
+class Solution_Iter {
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> res;
+        stack<TreeNode*> st;
+        TreeNode *curr = root;
+        while (curr != NULL || !st.empty()) {
+            if (curr != NULL) {
+                st.push(curr);
+                curr = curr->left;
+            }
+            else {
+                curr = st.top();
+                st.pop();
+                res.push_back(curr->val);
+                curr = curr->right;
+            }
+        }
+
+        return res;
+    }
+};
+
+// Morris traversal
+// time: O(n) - find prev pointer, space: O(1)
+class Solution_Morris {
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> res;
+        TreeNode *curr = root, *prev = NULL;
+        while (curr != NULL) {
+            if (curr->left == NULL) {
+                res.push_back(curr->val);
+                curr = curr->right;
+            }
+            else {
+                prev = curr->left;
+                while (prev->right != NULL && prev->right != curr) {
+                    prev = prev->right;
+                }
+
+                if (prev->right == NULL) {
+                    prev->right = curr;     // change tree shape
+                    curr = curr->left;
+                }
+                else {  // prev->right == curr
+                    prev->right = NULL;    // recover tree shape
+                    res.push_back(curr->val);
+                    curr = curr->right;
+                }
+            }
+        }
+
+        return res;
+    }
+};
+
+int main() {
+    TreeNode node3(3), node2(2, &node3, NULL), node1(1, NULL, &node2);
+    Solution sol;
+    vector<int> res = sol.inorderTraversal(&node1);
+    for (int i = 0; i < res.size(); i++) {
+        cout << res[i] << " ";
+    }
+    cout << endl;
+    cin.get();
+    return 0;
+}