题解 | #二叉树的前序遍历#前中后思路一致，举一反三递归+非递归

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型一维数组
     */
    public int[] preorderTraversal (TreeNode root) {
        // write code here
        //第一步 用list保存前序遍历结果
        //第二步 转为int[] 返回
        ArrayList<Integer> list = new ArrayList<>();
        pre(root,list);
        int[] preArray = new int[list.size()];
        for(int i = 0;i<list.size();i++){
            preArray[i] = list.get(i);
        }
            return preArray;
    }
    
    private void pre(TreeNode root,ArrayList<Integer> list){
        if(root == null){
            return;
        }
        //前中后只是调整下面三行的顺序即可
        list.add(root.val);
        pre(root.left,list);
        pre(root.right,list);
    }
}

// =============== 非递归方式 =====================
    public static void preOrder(Node head,ArrayList<Integer> list){
        if(head == null){
            return;
        }
        System.out.println("先序遍历，非递归方式");
        Stack<Node> s = new Stack<Node>();
        s.push(head);
        while( !s.isEmpty() ){
            head = s.pop();
            list.add(head.val);
            if(head.right != null){
                s.push(head.right);
            }
            if(head.left != null){
                s.push(head.left);
            }
        }
    }