144. 二叉樹的前序遍歷

難度：簡單

題目

給你二叉樹的根節點 root ，返回它節點值的 前序 遍歷。

示例 1：

輸入：root = [1,null,2,3]
輸出：[1,2,3]

示例 2：

輸入：root = []
輸出：[]

示例 3：

輸入：root = [1]
輸出：[1]

示例 4：

輸入：root = [1,2]
輸出：[1,2]

示例 5：

輸入：root = [1,null,2]
輸出：[1,2]

提示：

• 樹中節點數目在范圍 [0, 100] 內
• -100 <= Node.val <= 100

**進階：**遞歸算法很簡單，你可以通過迭代算法完成嗎？

個人題解

方法一：遞歸

遞歸，最簡單的前中后序遍歷方式

/*** 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<>();process(result, root);return result;}private void process(List<Integer> result, TreeNode node) {if (node == null) {return;}result.add(node.val);process(result, node.left);process(result, node.right);}
}

方法二：迭代

/*** 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<>();if (root == null) {return result;}Deque<TreeNode> stack = new LinkedList<>();stack.push(root);while (!stack.isEmpty()) {TreeNode cur = stack.pop();result.add(cur.val);if (cur.right != null) {stack.push(cur.right);}if (cur.left != null) {stack.push(cur.left);}}return result;}
}