题目:

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?

题解:

递归做法如下:

 1     public void helper(TreeNode root, ArrayList<Integer> re){
 2         if(root==null)
 3             return;
 4         re.add(root.val);
 5         helper(root.left,re);
 6         helper(root.right,re);
 7     }
 8     public ArrayList<Integer> preorderTraversal(TreeNode root) {
 9         ArrayList<Integer> re = new ArrayList<Integer>();
         if(root==null)
             return re;
         helper(root,re);
         return re;
     }

非递归方法:

 1 public ArrayList<Integer> preorderTraversal(TreeNode root) {
 2     ArrayList<Integer> res = new ArrayList<Integer>();
 3     if(root == null)
 4         return res;
 5     LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
 6     while(root!=null || !stack.isEmpty()){
 7         if(root!=null){
 8             stack.push(root);
 9             res.add(root.val);
             root = root.left;
         }
         else{
             root = stack.pop();
             root = root.right;
         }
     }
     return res;
 }

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