Binary-tree-preorder-traversal

题目描述

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 you do it iteratively?

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    ArrayList<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        if(root == null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.isEmpty())
        {
            TreeNode te = stack.pop();
            list.add(te.val);
            if(te.right != null)
            {
                stack.push(te.right);
            }
            if(te.left != null)
            {
                stack.push(te.left);
            }
        }
        return list;
    }
}

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
//递归，先序遍历：根 左 右
import java.util.*;
public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if(root == null){           //判断root==null
            return list;            //{}返回为[]
        }
        list.add(root.val);
        list.addAll(preorderTraversal(root.left));      //addAll:添加集合
        list.addAll(preorderTraversal(root.right));
        return list;
    }
}