[leedcode 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 {
    //第一个是非递归解法：
    //借助栈数据结构，将最左侧的节点入栈，到达最左侧节点时，弹出栈顶，填入结果，然后遍历栈顶的右节点
    //第二种是递归，比较简单
   /* public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res=new ArrayList<Integer>();
        Stack<TreeNode> stack=new Stack<TreeNode>();
        TreeNode node=root;
        while(!stack.empty()||node!=null){
            while(node!=null){

                stack.push(node);
                node=node.left;
            }
            node=stack.pop();
            res.add(node.val);
            node=node.right;
        }
        return res;
    }*/
    List<Integer> res=new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {
        getInorder(root);
        return res;
    }
    public void getInorder(TreeNode root){
          if(root==null) return;
          inorderTraversal(root.left);
          res.add(root.val);
          inorderTraversal(root.right);
    }
}