给定一个二叉树的根节点 root
,返回 它的 中序 遍历 。
输入:root = [1,null,2,3]
输出:[1,3,2]
输入:root = []
输出:[]
输入:root = [1]
输出:[1]
中序遍历:
左、根、右
/**
* 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 static List<Integer> inorderTraversal(TreeNode root) {
LinkedList<Integer> list = new LinkedList<>();
inorder(root, list);
return list;
}
public static void inorder(TreeNode root, LinkedList<Integer> list) {
if (root==null) {
return;
}
inorder(root.left, list);
list.add(root.val);
inorder(root.right, list);
}
}
执行用时:0 ms, 在所有 Java 提交中击败了100.00%的用户
内存消耗:39.5 MB, 在所有 Java 提交中击败了74.54%的用户
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