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[LeetCode] Binary Tree Postorder Traversal

时间:2015-06-03 00:54:17      阅读:101      评论:0      收藏:0      [点我收藏+]

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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 vector<int> postorderTraversal(TreeNode* root) {
 2     vector<int> nodes;
 3     stack<TreeNode*> toVisit;
 4     TreeNode* curNode = root;
 5     TreeNode* lastNode = NULL;
 6     while (curNode || !toVisit.empty()) {
 7         if (curNode) {
 8             toVisit.push(curNode);
 9             curNode = curNode -> left;
10         }
11         else {
12             TreeNode* topNode = toVisit.top();
13             if (topNode -> right && lastNode != topNode -> right)
14                 curNode = topNode -> right;
15             else {
16                 nodes.push_back(topNode -> val);
17                 lastNode = topNode;
18                 toVisit.pop();
19             }
20         }
21     }
22     return nodes;
23 }

Recursive solution:

 1 void postorder(TreeNode* root, vector<int>& nodes) {
 2     if (!root) return; 
 3     postorder(root -> left, nodes);
 4     postorder(root -> right, nodes);
 5     nodes.push_back(root -> val);
 6 }
 7 vector<int> postorderTraversal(TreeNode* root) {
 8     vector<int> nodes;
 9     postorder(root, nodes);
10     return nodes;
11 } 

Morris traversal:

 1 void reverseNodes(TreeNode* start, TreeNode* end) {
 2     if (start == end) return;
 3     TreeNode* x = start;
 4     TreeNode* y = start -> right;
 5     TreeNode* z;
 6     while (x != end) {
 7         z = y -> right;
 8         y -> right = x;
 9         x = y;
10         y = z;
11     }
12 }
13 void reverseAddNodes(TreeNode* start, TreeNode* end, vector<int>& nodes) {
14     reverseNodes(start, end);
15     TreeNode* node = end;
16     while (true) {
17         nodes.push_back(node -> val);
18         if (node == start) break;
19         node = node -> right;
20     }
21     reverseNodes(end, start);
22 }
23 vector<int> postorderTraversal(TreeNode* root) {
24     vector<int> nodes;
25     TreeNode* dump = new TreeNode(0);
26     dump -> left = root;
27     TreeNode* curNode = dump;
28     while (curNode) {
29         if (curNode -> left) {
30             TreeNode* predecessor = curNode -> left;
31             while (predecessor -> right && predecessor -> right != curNode)
32                 predecessor = predecessor -> right;
33             if (!(predecessor -> right)) {
34                 predecessor -> right = curNode;
35                 curNode = curNode -> left;
36             }
37             else {
38                 reverseAddNodes(curNode -> left, predecessor, nodes);
39                 predecessor -> right = NULL;
40                 curNode = curNode -> right;
41             }
42         }
43         else curNode = curNode -> right;
44     }
45     return nodes;
46 }

 

[LeetCode] Binary Tree Postorder Traversal

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原文地址:http://www.cnblogs.com/jcliBlogger/p/4548024.html

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