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難度：容易

要求：

給出一棵二叉樹，返回其節點值的前序遍歷。
樣例給出一棵二叉樹
{1,#,2,3}

 1 
   \ 
    2 
   / 
 3

返回 [1,2,3]

思路：非遞歸

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();  
        TreeNode node = root;  
        while(node != null || stack.size() > 0){  
            if(node != null){
                list.add(node.val);//壓棧之前先訪問  
                stack.push(node);//將所有左孩子壓棧
                node = node.left;  
            }else{  
                node = stack.pop();  
                node = node.right;  
            }  
        }  
        return list;
    }
}

思路：遞歸

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        traverse(root, result);
        return result;
    }
    // 把root為跟的preorder加入result里面
    private void traverse(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
        }
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}