[Leetcode] Binary Tree postorder Traversal

Given a binary tree, return the postorder traversal of its nodes ' values.

For example:
Given binary Tree {1,#,2,3},

return [3,2,1].

Note:recursive solution is trivial, could do it iteratively?

Recursive implementation code

/***************************************************************** * @Author: Chu hing * @Date: 2015/2/24 01 : * @Status: Accepted * @Runtime: 4 ms******************************************************************/
     structtreenode{intVal    TreeNode *left;    TreeNode *right; TreeNode (intx): Val (x), left (null), right (null) {}};classSolution { Public:voidHelper (TreeNode *root, vector<int>&result) {if(Root)            {Helper (root->left, result);            Helper (root->right, result);        Result.push_back (Root->val); }    } vector<int>Postordertraversal (TreeNode *root) { vector<int>Result Helper (root, result);returnResult }};

Non-recursive implementation code 1

classSolution { Public: vector<int>Postordertraversal (TreeNode *root) { vector<int>Resultif(Root = NULL) {returnResult } list<TreeNode*>Nodes        TreeNode *p = root;        TreeNode *cur; Nodes.push_back (P); while(!nodes.empty ()) {cur = Nodes.front ();//① The current node is not a leaf node, but the child has traversed it.             //② The current node is the leaf node.            if(cur->right = = P | | cur->left = = P | |            (Cur->left = = NULL && Cur->right = = null))                {Result.push_back (cur->val);                Nodes.pop_front ();            p = cur; }Else{if(Cur->right! = NULL)                {Nodes.push_front (cur->right); }if(Cur->left! = NULL)                {Nodes.push_front (cur->left); }            }        }returnResult }};

Non-recursive implementation code 2

classSolution { Public: vector<int>Postordertraversal (TreeNode *root) { vector<int>Result Stack<TreeNode*>Nodes        TreeNode *p = NULL; TreeNode *cur = root; while(cur! = NULL | |!nodes.empty ()) {if(cur! = NULL) { while(cur! = NULL)                    {Nodes.push (cur);                Cur = cur->left; }            }Else{cur = nodes.top ();if(cur->right = = NULL | | cur->right = = p)                    {Nodes.pop ();                    p = cur;                    Result.push_back (Cur->val);                cur = NULL; }Else{cur = cur->right; }            }        }returnResult }};

Implementation of the first order traversal see blog binary Tree preorder traversal

[Leetcode] Binary Tree postorder Traversal