Given binary tree {1,#,2,3},
1
2
/
3
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
Solution: 1. Iterative way (stack). Time: O(n), Space: O(n).
2. Recursive solution. Time: O(n), Space: O(n).
3. Threaded tree (Morris). Time: O(n), Space: O(1).
You may refer to my blog for more detailed explanations:
http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html
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> res; 14 postorderTraversalRe(root, res); 15 return res; 16 } 17 18 void postorderTraversalRe(TreeNode* root, vector<int> &res) 19 { 20 if(!root) return; 21 postorderTraversalRe(root->left, res); 22 postorderTraversalRe(root->right, res); 23 res.push_back(root->val); 24 } 25 };