Add Solution for 145 Binary Tree Postorder Traversal

cpp/145_Binary_Tree_Postorder_Traversal.cpp

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+// 145. Binary Tree Postorder Traversal
+/**
+ * Given a binary tree, return the postorder traversal of its nodes' values.
+ * 
+ * For example:
+ * Given binary tree {1,#,2,3},
+ *    1
+ *     \
+ *      2
+ *     /
+ *    3
+ * return [3,2,1].
+ * 
+ * Note: Recursive solution is trivial, could you do it iteratively?
+ * 
+ * Subscribe to see which companies asked this question.
+ * 
+ * Tags: Tree Stack
+ * 
+ * Similar Problems: (M) Binary Tree Inorder Traversal
+ * 
+ * Author: Kuang Qin
+ */
+
+#include <iostream>
+#include <vector>
+#include <stack>
+#include <algorithm>
+
+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)
+class Solution {
+    void postorder(TreeNode* root, vector<int>& nodes) {
+        if (root == NULL) {
+            return;
+        }
+
+        postorder(root->left, nodes);
+        postorder(root->right, nodes);
+        nodes.push_back(root->val);
+        return;
+    }
+public:
+    vector<int> postorderTraversal(TreeNode* root) {
+        vector<int> res;
+        postorder(root, res);
+        return res;
+    }
+};
+
+// interative solution using stack
+// time: O(n), space: O(n)
+class Solution_Iter {
+public:
+    vector<int> postorderTraversal(TreeNode* root) {
+        vector<int> res;
+        stack<TreeNode*> st;
+        TreeNode *curr = root;
+
+        // the opposite way of preorder
+        while (curr != NULL || !st.empty()) {
+            if (curr != NULL) {
+                st.push(curr);
+                res.push_back(curr->val);
+                curr = curr->right;
+            }
+            else {
+                curr = st.top();
+                st.pop();
+                curr = curr->left;
+            }
+        }
+
+        reverse(res.begin(), res.end());
+        return res;
+    }
+};
+
+// Morris Traversal
+// time: O(n), space: O(1)
+class Solution_Morris {
+    // reverse tree node from start to end
+    // 1
+    //  \               start: 1, end: 3
+    //   2
+    //    \             output: 3, 2, 1
+    //     3
+    void reverseNodes(TreeNode *start, TreeNode *end) {
+        if (start == end) {
+            return;
+        }
+
+        // can be treated as reversing a linked list, right is equal to next
+        TreeNode *prev = start, *curr = start->right, *next;
+        while (prev != end) {
+            next = curr->right; // save the next
+            curr->right = prev; // point to the start of the reversed list
+            prev = curr;        // update the start of the reversed list
+            curr = next;        // update the next node to be reversed
+        }
+
+        return;
+    }
+
+    void reverseAddNodes(TreeNode *start, TreeNode *end, vector<int>& nodes) {
+        reverseNodes(start, end);   // reverse for output
+
+        for (TreeNode *p = end; p != start; p = p->right) {
+            nodes.push_back(p->val);
+        }
+        nodes.push_back(start->val);
+
+        reverseNodes(end, start);   // recover the tree
+        return;
+    }
+public:
+    vector<int> postorderTraversal(TreeNode* root) {
+        vector<int> res;
+        TreeNode dummy(0), *curr = &dummy, *prev = NULL;
+        dummy.left = root;
+
+        while (curr != NULL) {
+            if (curr->left == NULL) {
+                curr = curr->right;
+            }
+            else {
+                prev = curr->left;
+                while (prev->right != NULL && prev->right != curr) {
+                    prev = prev->right; // find in-order previous node
+                }
+
+                if (prev->right == NULL) {
+                    prev->right = curr;
+                    curr = curr->left;
+                }
+                else {  // prev->right == curr
+                    // reverse add the right boundary of its left sub-tree
+                    reverseAddNodes(curr->left, prev, res);
+                    prev->right = NULL;
+                    curr = curr->right;
+                }
+            }
+        }
+
+        return res;
+    }
+};
+
+int main() {
+    TreeNode node3(3), node2(2, &node3, NULL), node1(1, NULL, &node2);
+    Solution sol;
+    vector<int> res = sol.postorderTraversal(&node1);
+    for (int i = 0; i < res.size(); i++) {
+        cout << res[i] << " ";
+    }
+    cout << endl;
+    cin.get();
+    return 0;
+}