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?

 

 

求前序遍历,要求不用递归。
 
 
使用双向队列。
 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        
        Deque<TreeNode> queue = new ArrayDeque<TreeNode>();

        List list = new ArrayList<Integer>();

        if( root == null )
            return list;

        queue.add(root);

        while( !queue.isEmpty() ){

            TreeNode node = queue.poll();
            list.add(node.val);

            if( node.right != null )
                queue.addFirst(node.right);
            if( node.left != null )
                queue.addFirst(node.left);


        }


        return list;

        
    }
}