Question

如何在O(n)上不停地转播O(n)的双read树,而不使用一ack(允许在温变数方面使用固定的额外空间,因此,我们可以在树中的每 no子上添加旗帜)。我花了很多时间思考这个问题,但我看不到什么东西可以做unless<>em>。我们正在向拥有树木数据的记忆点进行渗透。让我们说,我们正在利用多种阵列代表来执行点人,然后,我们可以tra笑O(n)的树,谁会想到什么东西?

<><>>>>> http://strong>not homework, only to Save theenergy of some key板 strokes to written comments about homework!

Answer 1

Let s say that we have the following threaded tree representation in C:

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<条码>O(1)上检索。

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
", 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:我没有这样做。

Answer 2

这是在 Java撰写的法典:

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
            }
        }
    }
}

纽埃语是一种内装饰语树木:

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;
    }
}

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