二叉树空间复杂度为O(1)的中序遍历c++

二叉树节点

typedef class BinaryTree
{
public:
    char ch;
    BinaryTree *parent;
    BinaryTree *LeftChild;
    BinaryTree *RightChild;
} * BT;

空间复杂度为O(1)的中序遍历

BT last(BT root) //寻找以root为根的最左叶子节点
{
    while (root->LeftChild != NULL)
    {
        root = root->LeftChild;
    }
    return root;
}

BT next(BT root) //返回按中序遍历规则的下一个节点
{
    if (root->RightChild != NULL)
        return last(root->RightChild);
    else
    {
        BT pa = root->parent;
        while (root != pa->RightChild && pa != NULL)
        {
            root = pa;
            pa = root->parent;
        }
        return pa;
    }
}

void inOrder(BT root)
{
    if (root == NULL)
        return;
    BT p = last(root);
    while (p != NULL)
    {
        cout << p->ch << " ";
        p = next(p);
    }
}