114 lines
2.2 KiB
Markdown
114 lines
2.2 KiB
Markdown
# 二叉树的中序遍历
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## 题目
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给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
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示例 1：
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输入：root = [1,null,2,3]
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输出：[1,3,2]
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示例 2：
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输入：root = []
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输出：[]
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示例 3：
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输入：root = [1]
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输出：[1]
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提示：
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树中节点数目在范围 [0, 100] 内
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-100 <= Node.val <= 100
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进阶: 递归算法很简单，你可以通过迭代算法完成吗？
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## 思路
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- 递归
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- 迭代
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## 代码
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- 递归
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public List<Integer> inorderTraversal(TreeNode root) {
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// Init: Res
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List<Integer> res = new ArrayList<>();
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inorder(root, res);
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return res;
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}
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public void inorder(TreeNode root, List<Integer> res) {
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if (root == null) {
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return;
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}
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inorder(root.left, res);
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res.add(root.val);
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inorder(root.right, res);
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}
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}
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```
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- 迭代
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public List<Integer> inorderTraversal(TreeNode root) {
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// Init: stack, res
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Deque<TreeNode> stack = new LinkedList<>();
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List<Integer> res = new ArrayList<>();
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TreeNode cur = root;
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// Traverse 1: Judge
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while (cur != null || stack.size() > 0) {
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// Traverse 2: Add
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while (cur != null) {
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stack.push(cur);
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cur = cur.left;
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}
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cur = stack.peek();
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stack.pop();
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res.add(cur.val);
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cur = cur.right;
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}
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return res;
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}
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}
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``` |