【LeetCode】94. 二叉树的中序遍历
94. 二叉树的中序遍历
难度：简单
题目
给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
示例 1：
输入：root = [1,null,2,3]
输出：[1,3,2]
示例 2：
输入：root = []
输出：[]
示例 3：
输入：root = [1]
输出：[1]
提示：
- 树中节点数目在范围
[0, 100]内 -100 <= Node.val <= 100
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
个人题解
方法一：递归
递归，最简单的前中后序遍历方式
/*** 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> inorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<>();process(result, root);return result;}public void process(List<Integer> result, TreeNode node) {if (node == null) {return;}process(result, node.left);result.add(node.val);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> inorderTraversal(TreeNode root) {List<Integer> res = new ArrayList<>();Deque<TreeNode> stack = new LinkedList<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()) {while (cur != null) {stack.push(cur);cur = cur.left;}cur = stack.pop();res.add(cur.val);cur = cur.right;}return res;}
}