Leetcode Binary Tree Preorder traversal

1. Topics

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

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

   1         2    /   3

Return [1,2,3] .

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

2. Solution 1

Class Solution {public:     vector<int> preordertraversal (TreeNode *root) {   & nbsp;    vector<int> ans;         deque<treenode*> node_list;         if (root = NULL) return ans;         Node_list.push_front (root);         while (!node_list.empty ())         {             TreeNode *cur = Node_list.back ();             Node_list.pop_back ();             ans.push_back (cur, Val);             if (cur-right! = NULL) Node_list.push_back ( cur, right);             if (cur-leFT! = NULL) node_list.push_back (cur, left);        }                   return ans;    };

idea: The non-recursive way of First order traversal is also relatively easy to write, that is, breadth first or breathing traversal. Need a queue to support.

3. Solution 2

Class Solution {public:    vector<int> path;    void preorder (TreeNode *root) {        if (!root)           return;        Path.push_back (root->val);        if (root->left)          preorder (root->left);        if (root->right)          preorder (root->right);    }    Vector<int> preordertraversal (TreeNode *root) {        preorder (root);        return path;    };

idea: Recursion is simple, but very slow.

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Leetcode Binary Tree Preorder traversal