94. Binary Tree Inorder Traversal

    /*
     * 94. Binary Tree Inorder Traversal 
     * 2015.11.21 By Mingyang
     * 跟preorder一样的dfs，不过是每次pop出来的时候记在小本本上，也就是记在list里面 中序遍历迭代解法
     * 用栈先把根节点的所有左孩子都添加到栈内， 然后输出栈顶元素，再处理栈顶元素的右子树
     */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if (root == null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode p = root;
        while (!stack.isEmpty() || p != null) {
            if (p != null) {
                stack.push(p);
                p = p.left;
            } else {
                TreeNode t = stack.pop();
                list.add(t.val);
                p = t.right;
            }
        }
        return list;
    }
    public List<Integer> inorderTraversal1(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        dfs(root, res);
        return res;
    }
    public void dfs(TreeNode root, List<Integer> res) {
        if (root == null)
            return;
        dfs(root.left, res);
        res.add(root.val);
        dfs(root.right, res);
    }