Question

如何在不使用堆栈的情况下在 O（n）中非递归地遍历线程化二叉树（只允许为临时变量使用常量额外空间，所以我们可以不要为树中的每个节点添加访问标志。我花了很多时间思考它，但在我看来似乎并不可行除非我们将遍历具有树数据的内存位置。假设我们使用多个数组表示来实现指针，那么我们可以在O（n）中遍历树，是否还有人想到其他东西？

注意这是不作业，只是为了节省一些键盘敲击的能量来写关于家庭作业的评论！

Answer 1

假设我们在C中有以下线程树表示：

typedef struct threaded_binary_tree {
    int value;

    // Flag that tells whether right points to a right child or to a next
    // node in inorder.
    bool right_is_child;

    struct threaded_binary_tree *left, *right;
} threaded_binary_tree;

然后，在O(1)内存中遍历它可能如下所示：

void inorder(threaded_binary_tree *node)
{
    threaded_binary_tree *prev = NULL;

    // Ignore empty trees
    if (node == NULL)
        return;

    // First, go to the leftmost leaf of the tree
    while (node->left != NULL)
        node = node->left;

    while (node != NULL) {
        // Nodes are visited along going upward in the tree
        printf("%d\n", node->value);

        prev = node;
        node = node->right;

        if (prev->right_is_child) {
            // We're descending into tree

            if (node != NULL) {
                // Again, go to the leftmost leaf in this subtree
                while (node->left != NULL)
                    node = node->left;
            }
        }
        // else, we're climbing up in the tree
    }
}

警告：我没有运行此代码。

Answer 2

这是用Java编写的代码：

public void inOrder() {
    Node<T> curr = root;
    boolean visited = false; //I haven't come across the node from which I came

    while (curr != null) {
        if (!visited && curr.left != null) { //Go to leftmost node
            curr = curr.left;
        } else {
            System.out.println(curr.head + " ");
            if (curr.right != null) { //I prioritize having childs than "threaded sons"
                curr = curr.right;
                visited = false;
            } else {
                curr = curr.rightThreaded;
                visited = true; //Means that I will come back to a node I've already looped, but now i'll print it, except if i'm at the last node
            }
        }
    }
}

Node是ThreadedBinaryTree的内部类：

private static class Node<T> {
    private T head;
    private Node<T> left;
    private Node<T> right;
    private Node<T> leftThreaded;
    private Node<T> rightThreaded;

    public Node(T head, Node<T> leftThreaded, Node<T> rightThreaded) {
        this.head = head;
        this.leftThreaded = leftThreaded;
        this.rightThreaded = rightThreaded;
    }
}

迭代螺纹二叉树遍历只有恒定的额外空间

2 个答案: