将二叉树上色为红黑树

常见的采访问题是给出一种算法，以确定给定的二叉树是否高度平衡（AVL树定义）。

我想知道我们是否可以对红黑树做类似的事情。

给定一个任意的未着色的二叉树（具有NULL节点），是否存在一种“快速”算法，该算法可以确定我们是否可以对节点Red / Black进行着色（并查找着色），以便它们满足Red-Black树的所有属性（这个问题的定义）？

最初的想法是，我们可以删除NULL节点，然后尝试递归地验证生成的树是否可以是一棵红黑树，但这似乎没有任何用处。

我在网上做了简短的搜索，但是似乎找不到任何可以解决这个问题的文件。

我可能错过了一些简单的东西。

— Aryabhata
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我敢肯定，一棵树可以是红色，黑色当且仅当每个节点，从中空节点的路径最长不超过两倍长于最短的一个。这样够快吗？

— KarolisJuodelė2013年

Answers:

如果对于一棵树的每个节点，从树到叶节点的最长路径比最短的树长不超过两倍，则该树具有红黑颜色。

这是一种计算任何节点颜色的算法 n

if n is root,
    n.color = black
    n.black-quota = height n / 2, rounded up.

else if n.parent is red,
    n.color = black
    n.black-quota = n.parent.black-quota.

else (n.parent is black)
    if n.min-height < n.parent.black-quota, then
        error "shortest path was too short"
    else if n.min-height = n.parent.black-quota then
        n.color = black
    else (n.min-height > n.parent.black-quota)
        n.color = red
    either way,
        n.black-quota = n.parent.black-quota - 1

这n.black-quota是您希望看到的从节点到叶子的黑色节点数，n并且n.min-height是到最近叶子的距离。

为简便起见，令，和。 $b(n) =$ n.black-quota $h(n) =$ n.height $m(n) =$ n.min-height

定理：修正了一个二叉树。如果为每个节点，和节点， $T$ $n \in T$ $h(n) \leq 2m(n)$ $r = \text{root}(T)$ 则具有红黑色，从根到叶的每条路径上都有正好是黑色节点。 $b(r) \in [\frac{1}{2}h(r), m(r)]$ $T$ $b(r)$

证明：对。 $b(n)$

验证高度为1或2的所有四棵树均满足定理。 $b(n) = 1$

根据红黑树的定义，根是黑色。令为具有黑色父级的节点，使得 $n$ $p$ 。然后，和。 $b(p) \in [\frac{1}{2}h(p), m(p)]$ $b(n) = b(p) -1$ $h(n) = h(p)-1$ $h(n) \geq m(n) \geq m(p)-1$

假设该定理对于所有根为树成立， $r$ 。 $b(r) < b(q)$

如果，则 $b(n) = m(n)$ $n$ 根据归纳假设可以为红黑色。

如果然后 $b(p) = \frac{1}{2}h(p)$ 。不满足归纳假设，因此必须为红色。令为的子代。和 $b(n) = \lceil \frac{1}{2}h(n) \rceil - 1$ $n$ $c$ $n$ $h(c) = h(p)-2$ 。然后 $b(c) = b(p)-1 = \frac{1}{2}h(p)-1 = \frac{1}{2}h(c)$ $c$ can be red-black colored by the inductive assumption.

Note that, by the same reasoning, if $b(n) \in (\frac{1}{2}h(r), m(r))$ , then both $n$ and a child of $n$ satisfy the inductive assumption. Therefore $n$ could have any color.

— Karolis Juodelė
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@Aryabhata, any traversal is fine, as long as the parent is seen before its children. I don't have a formal proof written, but I have an idea of how it would look. I'll try writing something up when I can.

— Karolis Juodelė

@Aryabhata, i added a proof. Sorry I took so long.

— Karolis Juodelė

@Aryabhata, the idea is that if

b (p)

$b(p)$ of some node

p

$p$ is withing certain bounds, a child or grandchild

c

$c$ of

p

$p$ can have

b (c)

$b(c)$ within those same bounds. Having

b (n)

$b(n)$ in those bounds may correspond to

n

$n$ being black. Most of the proof is about bounding

h

$h$ and

m

$m$ of a child or grandchild, given

h

$h$ and

m

$m$ of the parent or grandparent. Your tree is certainly colorable.

b (r o o t) = 8

$b(root) = 8$ , left child is black and right child is red, the path of length 16 is

b r b r b r \dots

$brbrbr\dots$ , the path of length 8 is

b b b b b b b b

$bbbbbbbb$ , paths of 9 and 12 can have multiple valid colorings.

— Karolis Juodelė

There's a question about this answer.

— David Richerby

I believe Karolis' answer is correct (and a pretty nice characterization of red-black trees, giving an $O(n)$ time algorithm), just wanted to add another possible answer.

One approach is to use dynamic programming.

Given a tree, for each node $n$ , you construct two sets: $S_R(n)$ and $S_B(n)$ which contains the possible black-heights for the subtree rooted at $n$ . $S_R(n)$ contains the black-heights assuming $n$ is coloured Red, and $S_B(n)$ assuming $n$ is coloured black.

Now given the sets for $n.Left$ and $n.Right$ (i.e direct children of $n$ ), we can compute the corresponding sets for $n$ , by taking appropriate intersections and unions (and incrementing as needed).

I believe this comes out be an $O(n \log n)$ time algorithm.

— Aryabhata
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