【LeetCode】144. 二叉树的前序遍历

题目描述：

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]
示例 2：

输入：root = []
输出：[]
示例 3：

输入：root = [1]
输出：[1]
示例 4：

输入：root = [1,2]
输出：[1,2]
示例 5：

输入：root = [1,null,2]
输出：[1,2]

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-preorder-traversal
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递归写法

/**
 * 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> arr = new ArrayList<Integer>();
        traverse(root, arr);
        return arr;
    }

    public void traverse(TreeNode root, List<Integer> list) {
    
    
        if (root == null) return;
        list.add(root.val);
        traverse(root.left, list);
        traverse(root.right, list);
    }
}

迭代写法

/**
 * 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> arr = new ArrayList<>();
        if (root == null) {
    
    
            return arr;
        }
        Stack<TreeNode> stack = new Stack<>();
        while (!stack.isEmpty() || root != null) {
    
    
            while (root != null) {
    
    
                arr.add(root.val);
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            root = root.right;
        }
        return arr;
    }
}