力扣hot100 - 144、二叉树的前序遍历

题目：

法一：递归

法二：迭代

整体思路：先将根节点放入栈中，弹出并放入结果集合，再将右、左孩子放入栈，弹出一个并且放入结果集合，将弹出的节点有左孩子放入栈中。为什么是先右后左？弹出时刚好是先左后右。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList();
        Deque<TreeNode> stack = new ArrayDeque();
        if(root == null) return result;
        stack.push(root);

        while(!stack.isEmpty()){
            TreeNode node = stack.peek();
            stack.pop();
            if(node != null){
                result.add(node.val);
                if(node.right != null){
                    stack.push(node.right);
                }
                if(node.left != null){
                     stack.push(node.left);
                }
            
            }else{
                continue;
            }
        }

        return result;
    }
}