[LeetCode] Binary Tree Postorder Traversal, leetcodepostorder

[LeetCode] Binary Tree Postorder Traversal, leetcodepostorder

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, cocould you do it iteratively?

Recursive Implementation Code

/*************************************** * ************************** @ Author: chuxing * @ Date: 2015/2/24 * @ Status: Accepted * @ Runtime: 4 ms ************************************** * **************************/struct TreeNode {int val; treeNode * left; TreeNode * right; TreeNode (int x): val (x), left (NULL), right (NULL) {}}; class Solution {public: void helper (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); return result ;}};

Non-Recursive Implementation Code 1

Class Solution {public: vector <int> postorderTraversal (TreeNode * root) {vector <int> result; if (root = NULL) {return result ;} 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 already 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) ;}}return result ;}};

Non-Recursive Implementation Code 2

class Solution {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;                }            }        }        return result;    }};

For the implementation of first-order Traversal, see the blog post Binary Tree Preorder Traversal.