遍历二叉树——无递归、无栈、无树修改

Question

因此，对于我正在构建的应用程序，我需要能够在不使用递归、堆栈或在创建树后以任何方式修改树的情况下遍历二叉树。我的节点结构如下：

typedef struct 
{
    ValueType value;  //Data stored in node
    int left_index;   //Left child
    int right_index;  //Right child
    int parent_index; //Parent node
}

我将树存储为一维数组，其中每个节点的左子节点位于索引 2*i + 1，右子节点位于 2*i + 2，父节点位于 [i-1]/2 .如果一个节点没有父节点或子节点，它的关联索引为-1。

我发现的唯一一种基于迭代的非堆栈算法是一种叫做“Morris Traversal”的算法，如下所示：http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/

但是，Morris Traversal 在遍历期间修改了树，这是我无法做到的。

我愿意为每个节点添加任何需要的信息，只要我能在上述约束条件下编写算法即可。

我所要求的是否可能？如果是这样，我将如何去做？甚至不确定如何开始。

Answer 1

“while(!done)”循环不够吗？

Answer 2

你想要的是threaded binary tree。所有叶子节点的右指针指向该节点的有序后继。

创建这样的东西很容易，在插入或删除节点时更新它一点也不难。如果您可以控制节点结构，那么这几乎肯定是您想要的。

Answer 3

public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode parent;

    public void traverse() {
        TreeNode current = this.leftMost();
        while (current != null) {
            System.out.println("Current at " + current.val);
            current = current.inOrderNext();
        }
    }

    public TreeNode inOrderNext() {
        if (right != null) {
            return right.leftMost();
        } else {
            TreeNode current = this;
            TreeNode above = this.parent;
            while (true) {
                if (above == null) {
                    return null;
                } else {
                    if (above.left == current) {
                        return above;
                    } else {
                        current = above;
                        above = above.parent;
                    }
                }
            }
        }
    }

    public TreeNode leftMost() {
        TreeNode result = this;
        while (result.left != null) {
            result = result.left;
        }
        return result;
    }

    public static void main(String args[]) {
        TreeNode first = new TreeNode();
        first.val = 4;

        TreeNode second = new TreeNode();
        second.val = 2;
        second.parent = first;
        first.left = second;

        TreeNode third = new TreeNode();
        third.val = 1;
        third.parent = second;
        second.left = third;

        third = new TreeNode();
        third.val = 3;
        third.parent = second;
        second.right = third;

        second = new TreeNode();
        second.val = 6;
        second.parent = first;
        first.right = second;

        third = new TreeNode();
        third.val = 5;
        third.parent = second;
        second.left = third;

        third = new TreeNode();
        third.val = 7;
        third.parent = second;
        second.right = third;

        first.traverse();
    }
}