Total Accepted: 17403 Total Submissions: 50093 My Submissions
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?
【解題思路】
1、遞迴

Java AC

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private ArrayList<Integer> list;
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        list = new ArrayList<Integer>();
        preOrder(root);
        return list;
    }
    private void preOrder(TreeNode root){
        if(root == null){
            return;
        }
        list.add(root.val);
        preOrder(root.left);
        preOrder(root.right);
    }
}
 

2、非遞迴 棧作為輔助，每次先放入右孩子，再放入左孩子。依次出棧就是前序遍歷。

Java AC

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