【每日一题】【集合非空判断】【逻辑运算符】2021年12月15日-144. 二叉树的前序遍历

/**
 * 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 {
    List<Integer> res = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        preOrder(root);
        return res;
    }

    public void preOrder(TreeNode root) {
        if(root == null) {
            return;
        }
        Stack<TreeNode> stack = new Stack<>();
        //表示栈非空需要使用函数
        while(!stack.isEmpty() || root != null) {
            while(root != null) { //栈内元素可能为空，只剩一个root，右子树
                res.add(root.val);
                stack.push(root);
                root = root.left;
            }
            root = stack.pop().right;
        }
    }
}