描述

给出一棵二叉树，返回其节点值的前序遍历。

样例

给出一棵二叉树 {1,#,2,3},

   1
    \
     2
    /
 3

返回 [1,2,3].

挑战

你能使用非递归实现么？

前序

根左右

代码

/**
 * 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;
 *     }
 * }
 */

1. 非递归＋栈(重点)

PS：  
   注意区分stack.isEmpty()和stack == null  

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        List<Integer> preorder = new ArrayList<>();
        
        if (root == null) {
            return preorder;
        }

        stack.push(root);
        while (!stack.empty()) {
            TreeNode node = stack.pop();
            preorder.add(node.val);
            if (node.right != null) {
                stack.push(node.right);
            }
            if (node.left != null) {
                stack.push(node.left);
            }
            // 注意压入栈的顺序，先右后左入栈出来才能先左后右
        }

        return preorder;
    }
}

1. 遍历+递归

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;
    }
        
    private void traverse(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
            // return;表示不做任何操作
        }
        
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}

1. 分治+递归

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>();
        if (root == null) {
            return result;
        }
           
        ArrayList left = preorderTraversal(root.left);
        ArrayList right = preorderTraversal(root.right);
           
        result.add(root.val);
        result.addAll(left);
        result.addAll(right);
        // 三行代码的添加顺序不能变，即前序遍历的根左右
        return result;
    }
}

关于ArrayList.addAll():

result.addAll(list); // 把list中的每一个元素加到result中，
result.size()==list.size()
result.add(list); // 将list作为一个元素加到result中，则
result.size()为1
在使用ArrayList的addAll()方法的时候一定要进行list非null的判断