144. Binary Tree Preorder Traversal

Given a binary tree, return thepreordertraversal of its nodes' values.

For example:
Given binary tree[1,null,2,3],

   1
    \
     2
    /
   3

return[1,2,3].

Thoughts:

Using Stack

Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode *> treeSt;
        vector<int> answer;
        if(!root) return vector<int>();

        treeSt.push(root);
        while(!treeSt.empty()){
            TreeNode* cur = treeSt.top();
            answer.push_back(cur->val);
            treeSt.pop();
            if(cur->right)treeSt.push(cur->right);
            if(cur->left)treeSt.push(cur->left);
        }

        return answer;
    }
};

Code 2: Check from the pop

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> answer;
        // if(!root) return answer;
        stack<TreeNode *> treeSt;
        treeSt.push(root);
        while(!treeSt.empty()){

            TreeNode* cur = treeSt.top();
            treeSt.pop();
            if (cur){
                answer.push_back(cur ->val);
                treeSt.push(cur->right);
                treeSt.push(cur->left);
            }

        }

        return answer;
    }
};

Code (Python)

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack, res = [root], []
        while stack:
            node = stack.pop()
            if node: 
                res.append(node.val)
                stack.append(node.right)
                stack.append(node.left)
        return res