LintCode -- 二叉树的前序遍历

LintCode -- binary-tree-preorder-traversal(二叉树的前序遍历)

原题链接：http://www.lintcode.com/zh-cn/problem/binary-tree-preorder-traversal/

给出一棵二叉树，返回其节点值的前序遍历。

您在真实的面试中是否遇到过这个题？  

Yes

样例

给出一棵二叉树 {1,#,2,3},

   1
    \
     2
    /
   3

 返回 [1,2,3].

挑战

你能使用非递归实现么？

非递归实现（C++、Java、Python）：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in vector which contains node values.
     */
    vector preorderTraversal(TreeNode *root) {
        // write your code here
        TreeNode *x = root;
        vector p;
        vector res;
        while(x != NULL || p.size() != 0){
            res.push_back(x->val);
            if (x->right != NULL)
                p.push_back(x->right);
            x = x->left;
            if (x == NULL && p.size() != 0){
                x = p.back();
                p.pop_back();
            }
        }
        return res;
    }
};

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList res = new ArrayList();
        ArrayList p = new ArrayList();
        while (root != null || p.size() != 0){
            res.add(root.val);
            if (root.right != null)
                p.add(root.right);
            root = root.left;
            if (root == null && p.size() != 0){
                root = p.get(p.size()-1);
                p.remove(p.size()-1);
            }
        }
        return res;
    }
}

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""


class Solution:
    """
    @param root: The root of binary tree.
    @return: Preorder in ArrayList which contains node values.
    """
    def preorderTraversal(self, root):
        # write your code here
        p = [root]
        res = [0]
        while root is not None or len(p) != 1:
            res.append(root.val)
            if root.right is not None:
                p.append(root.right)
            root = root.left
            if root == None and len(p) != 1:
                root = p[len(p) - 1]
                del p[len(p) - 1]
        n = len(res)
        return res[1:n]