LeetCode144 二叉树先序遍历

Given a binary tree, return the preorder traversal of its nodes’ values.

Example:

Input: [1,null,2,3]
1

2
/
3

Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
递归：

public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> pre = new LinkedList<Integer>();
		prehelper(root, pre);
		return pre;
    }
    private void prehelper(TreeNode root, List<Integer> pre){
        if(root == null) return;
        pre.add(root.val);
        prehelper(root.left, pre);
        prehelper(root.right, pre);
    }

迭代：

public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> pre = new LinkedList<Integer>();
        if(root == null) return pre;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            pre.add(node.val);
            if(node.right != null) stack.push(node.right);
            if(node.left != null) stack.push(node.left);
        }
        return pre;
    }