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?

答案

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result=null;
        if(root==null)
        {
            result=new LinkedList<Integer>();
            return result;
        }
        if(root.left==null&&root.right==null)
        {
            result=new LinkedList<Integer>();
            result.add(root.val);
            return result;
        }

        if(root.left!=null)
        {
            result=preorderTraversal(root.left);
        }
        if(root.right!=null)
        {
            List<Integer> right=preorderTraversal(root.right);
            if(result==null)
            {
                result=right;
            }
            else
            {
                result.addAll(right);
            }
        }
        result.add(0,root.val);
        return result;
    }
}