lintcode ：Binary Tree Preorder Traversal 二叉树的前序遍历

题目：

二叉树的前序遍历

给出一棵二叉树，返回其节点值的前序遍历。

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？

解题：

通过递归实现，根节点->左节点->右节点

Java程序：

/**
 * 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<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> res = new ArrayList<Integer>();  
        res = preTurn(res,root);
        return res;  
    }
    public ArrayList<Integer> preTurn(ArrayList<Integer> res ,TreeNode root){
        if(root==null)
            return res;
        if(root!=null){
            res.add(root.val);
            if(root.left!=null){
                res= preTurn(res,root.left);
            }
            if(root.right!=null){
                res = preTurn(res,root.right);
            }
        }
        return res;
    }
}

总耗时: 1094 ms

 Python程序：

"""
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
        res = []
        res = self.preorderTurn(res,root)
        return res
    
    def preorderTurn(self,res,root):
        if root==None:
            return res
        if root!=None:
            res.append(root.val)
        if root.left!=None:
            res = self.preorderTurn(res,root.left)
        if root.right!=None:
            res = self.preorderTurn(res,root.right)
        return res;

总耗时: 335 ms

非递归程序，直接来源

Java程序：

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<TreeNode> p = new ArrayList<TreeNode>();  
        ArrayList<Integer> res = new ArrayList<Integer>();  
        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;  
     
    }
}

总耗时: 1473 ms

Python程序：

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
        res = []
        p = [root]  
        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]  
        return res

总耗时: 230 ms