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？
解题答案
/**
* 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;
* }
* }
*/
// 栈
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
TreeNode node = root;
Stack<TreeNode> stack = new Stack<>();
while (!stack.isEmpty() || node != null) {
if (node != null) {
stack.push(node);
node = node.left;
} else {
TreeNode top = stack.pop();
res.add(top.val);
node = top.right;
}
}
return res;
}
}
/**
* 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;
* }
* }
*/
// 递归
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
inorder(root, res);
return res;
}
private void inorder(TreeNode node, List<Integer> list) {
if (node == null) return;
inorder(node.left, list);
list.add(node.val);
inorder(node.right, list);
}
}
