LeetCode – Refresh – Binary Tree Post Order Traversal

Still 3 methods:

Same with inorder, but post order is a little different. We need to treat it as reverse result of preorder.

So we go to right branch first, then go back to left branch

1. recursive (use memory stack);

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void getTree(TreeNode *root, vector<int> &result) {
13         if (!root) return;
14         getTree(root->left, result);
15         getTree(root->right, result);
16         result.push_back(root->val);
17     }
18     vector<int> postorderTraversal(TreeNode *root) {
19         vector<int> result;
20         getTree(root, result);
21         return result;
22     }
23 };
24     

2. directly use stack:

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode *root) {
13         vector<int> result;
14         stack<TreeNode *> s;
15         if (!root) return result;
16         while (root || !s.empty()) {
17             if (root) {
18                 result.push_back(root->val);
19                 s.push(root->left);
20                 root = root->right;
21             } else {
22                 root = s.top();
23                 s.pop();
24             }
25         }
26         for (int i = 0; i < result.size()/2; i++) {
27             [](int &a, int &b) {int t = a; a = b; b = t;}(result[i], result[result.size()-i-1]);
28         }
29         return result;
30     }
31 };
32     

3. two pointers:

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode *root) {
13         vector<int> result;
14         if (!root) return result;
15         TreeNode *prev = NULL;
16         while (root) {
17             result.push_back(root->val);
18             if (!root->right) {
19                 root = root->left;
20             } else {
21                 prev = root->right;
22                 while(prev->left && prev->left != root) prev = prev->left;
23                 if (!prev->left) {
24                     prev->left = root;
25                     root = root->right;
26                 } else {
27                     prev->left = NULL;
28                     root = root->left;
29                     result.pop_back();
30                 }
31             }
32         }
33         for (int i = 0; i < result.size()/2; i++) {
34             [](int &a, int &b) {int t = a; a = b; b = t;}(result[i], result[result.size()-i-1]);
35         }
36         return result;
37     }
38 };
39