Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,3,2]
解释下题目:
就是先遍历左子树,然后遍历根节点,最后遍历右子树。。递归简直不要太简单
1. 递归
实际耗时:0ms
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
inorder(root, res);
return res;
}
private void inorder(TreeNode cur, List<Integer> list) {
if (cur == null) {
return;
} else {
inorder(cur.left, list);
list.add(cur.val);
inorder(cur.right, list);
}
}
这还要思路?