仅具有恒定额外空间的迭代线程二叉树遍历答案

【问题标题】：Iterative Threaded Binary Tree Traversing with only constant extra space仅具有恒定额外空间的迭代线程二叉树遍历
【发布时间】：2012-05-06 18:59:07
【问题描述】：

如何在O(n)中非递归地遍历线程二叉树而不使用堆栈（只允许为临时变量使用恒定的额外空间，所以我们可以'不要将访问标志添加到树中的每个节点）。我花了很长时间思考这个问题，但在我看来这并不可行除非我们要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，有没有其他想法？

注意这不是作业，只是为了节省一些键盘敲击的能量来写关于作业的cmets！

【问题讨论】：

你看过Wikipedia吗？
@svick 算法的第二步是 Step-2：将左子节点放入已访问节点列表中，并将其设为当前节点。转到步骤 6。 我在问题中提到我需要使用 constant extra space
我的回答与您的问题相关吗？如果有，请将其标记为一个。如果不是，请说明原因，我会尽力完善它。

标签： data-structures binary-tree tree-traversal non-recursive

【解决方案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;

那么，在O(1)内存中遍历它可能是这样的：

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：我还没有运行此代码。

【讨论】：

【解决方案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;
    }
}

【讨论】：