LeetCode-94. Binary Tree Inorder Traversal

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]

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

public List<Integer> inorderTraversal(TreeNode root) {// 树的中序遍历 递归 my
        List<Integer> res = new ArrayList<Integer>();
        inorderT(root,res);
        return res;

    }
    private void inorderT(TreeNode root,List list){
        if(null!=root){
            inorderT(root.left,list);
            list.add(root.val);
            inorderT(root.right,list);
        }
    }

　　

public List<Integer> inorderTraversal(TreeNode root) {//树的中序遍历 迭代 myTip
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while(null!= cur || !stack.isEmpty()){
            while(null!=cur){
                stack.push(cur);
                cur=cur.left;
            }
            cur= stack.pop();
            res.add(cur.val);
            cur= cur.right;
        }
        return res;

    }