[LintCode] 二叉树的后序遍历

The recursive solution is trivial and I omit it here.

Iterative Solution using Stack (O(n) time and O(n) space):

 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14     /**
15      * @param root: The root of binary tree.
16      * @return: Postorder in vector which contains node values.
17      */
18 public:
19     vector<int> postorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> nodes;
22         TreeNode* node = root;
23         TreeNode* lastNode = NULL;
24         stack<TreeNode*> toVisit;
25         while (node || !toVisit.empty()) {
26             if (node) {
27                 toVisit.push(node);
28                 node = node -> left;
29             }
30             else {
31                 TreeNode* topNode = toVisit.top();
32                 if (topNode -> right && topNode -> right != lastNode)
33                     node = topNode -> right;
34                 else {
35                     nodes.push_back(topNode -> val);
36                     lastNode = topNode;
37                     toVisit.pop();
38                 }
39             }
40         }
41         return nodes;
42     }
43 };

Another sophisticated solution using Morris Traversal (O(n) time and O(1) space):

 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14     /**
15      * @param root: The root of binary tree.
16      * @return: Postorder in vector which contains node values.
17      */
18 public:
19     vector<int> postorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> nodes;
22         TreeNode* dump = new TreeNode(0);
23         dump -> left = root;
24         TreeNode* node = dump;
25         while (node) {
26             if (node -> left) {
27                 TreeNode* predecessor = node -> left;
28                 while (predecessor -> right && predecessor -> right != node)
29                     predecessor = predecessor -> right;
30                 if (!(predecessor -> right)) {
31                     predecessor -> right = node;
32                     node = node -> left;
33                 }
34                 else {
35                     reverseAddNodes(node -> left, predecessor, nodes);
36                     predecessor -> right = NULL;
37                     node = node -> right;
38                 }
39             }
40             else node = node -> right;
41         }
42         return nodes;
43     }
44 private:
45     void reverseNodes(TreeNode* start, TreeNode* end) {
46         TreeNode* x = start;
47         TreeNode* y = start -> right;
48         TreeNode* z;
49         while (x != end) {
50             z = y -> right;
51             y -> right = x;
52             x = y;
53             y = z;
54         }
55     }
56     void reverseAddNodes(TreeNode* start, TreeNode* end, vector<int>& nodes) {
57         reverseNodes(start, end);
58         TreeNode* node = end;
59         while (true) {
60             nodes.push_back(node -> val);
61             if (node == start) break;
62             node = node -> right;
63         }
64         reverseNodes(end, start);
65     }
66 };

 

posted @ 2015-06-29 17:08  jianchao-li  阅读(219)  评论(0)  编辑  收藏  举报