94.Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:

Given binary tree

{1,#,2,3}

,

1
\
2
/
3

return

[1,3,2]

.

Note: Recursive solution is trivial, could you do it iteratively?

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/*
* 采用中序遍历
* 如果左子节点的list不为空，则将返回后的左子节点的list追加到当前节点的list中
* 存入当前节点的val至当前节点的list中
* 如果右子节点的list不为空，则将返回后的右子节点的list追加到当前节点的list中
*/
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
List<Integer> left;
List<Integer> right;

if(root == null)
return list;

left = inorderTraversal(root.left);//左
if(left.size() != 0){
for(int i = 0; i < left.size(); i++){
list.add(left.get(i));
}
}

list.add(root.val);//中

right = inorderTraversal(root.right);//右
if(right.size() != 0){
for(int j = 0; j < right.size(); j++){
list.add(right.get(j));
}
}

return list;
}
}