13 Red-black Trees
Red-black trees are one of many search-tree schemes that are "balanced" in order to guarantee that basic dynamic-set operations take O(lgn) time in the worst case.
Red-black trees 是许多搜索树框架中得一个。这些树为了保证基本的动态集合在最坏情况下操作时间在0(lgn )，采取了自平衡。
A red-black tree is a binary tree that satisfies the following red-black properties:
1. Every node is either red or black.
2. The root is black.
3. Every leaf (NIL) is black.
4. If a node is red, then both its children are black.
5. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.
1 所有的节点非红即黑
2 根是黑的
3 叶子（Nil）也是黑得
4 一个节点是红得，孩子必定是黑的。
5 对于每个节点，从这个节点到叶子的任意路径包含同样数量的黑的。
red-black tree 的属性从2-5感觉都是针对黑色的限制。
an example of a red-black tree.
As a matter of convenience in dealing with boundary conditions in red-black
tree code, we use a single sentinel to represent NIL
All pointers to NIL are replaced by pointers to the sentinel T.nil
所有指向空的都被替换成指向哨兵 T.nil 了
In the remainder of this chapter, we omit the leaves when we draw red-black trees, as shown
We call the number of black nodes on any simple path from, but not including, a node x down to a leaf the black-height of the node, denoted bh(x)
13.2 Rotations
We change the pointer structure through rotation, which is a local operation in a search tree that preserves the binary-search-tree property.
我们通过rotation来改变指针结构。它是保留搜索树属性的一个本地操作。
13.3 Insertion
We can insert a node into an n-node red-black tree in O(lgn) time.
To do so, we use a slightly modified version of the TREE-INSERT procedure (Section 12.3) to insert node ́ into the tree T as if it were an ordinary binary search tree, and then we color z red.
Case1:
Case 2 and case 3
13.4 Deletion
Like the other basic operations on an n-node red-black tree, deletion of a node takes time O(lgn ). Deleting a node from a red-black tree is a bit more complicated than inserting a node .
First ,we need to customize the TRANSPLANT subroutine that Tree-Delete calls so that it applies to a red-black tree :
Here is the red-black delete tree program :
Finally, if node y was black, we might have introduced one or more violations of the red-black properties, and so we call RB-DELETE-FIXUP in line 22 to restore the red-black properties. If y was red, the red-black properties still hold when y is removed or moved, for the following reasons:
如果节点y是黑色的，我们可能引入一个或多个违反红黑树性质。如果y是红色的，则红黑树的性质能得到保证。原因如下：
1. No black-heights in the tree have changed.
黑色深度没有变化
2. No red nodes have been made adjacent. Because y takes z's place in the tree, along with z's color, we cannot have two adjacent red nodes at y's new position in the tree. In addition, if y was not ́'s right child, then y's original right child x replaces y in the tree. If y is red, then x must be black, and so replacing y by x cannot cause two red nodes to become adjacent.
没有红色节点相邻。因为y取代了z得位置，并且取得了z得颜色，z原来是具有红黑树属性的，所以替换了以后仍然有。
另外，如果y不是z的右孩子，则y的原来位置被x取代了。如果y是红色的话，那么x肯定是黑色的，所以被x取代y 不可能导致两个红色节点相邻。
3. Since y could not have been the root if it was red, the root remains black.
如果y是红色的话，y肯定不是根，因此根仍然保持黑色。
If node y was black, three problems may arise, which the call of RB-DELETE- FIXUP will remedy.
First, if y had been the root and a red child of y becomes the new root, we have violated property 2.
首先如果y是根，并且y的一个红孩子成为了新根，那么违反 根是黑色的 属性。
Second, if both x and x.p are red, then we have violated property 4.
如果x和x.p 是红色的，那么我们可能违反 红色节点不能相邻这一条。
Third, moving y within the tree causes any simple path that previously contained y to have one fewer black node. Thus, property 5 is now violated by any ancestor of y in the tree.
第三，如果移动y，那么任意一条原先包含y的路线可能比其他的路线少一条黑色，因此，从任意一节点到叶子的黑色节点数是相同的。
We can correct the violation of property 5 by saying that node x, now occupying y's original position, has an "extra" black.
我们可以纠正 第五条属性 通过将现在占据y的原来位置的x的属性 有一个额外的黑色。
That is, if we add 1 to the count of black nodes on any simple path that contains x, then under this interpretation, property 5 holds.
也就是说，在我们计算从任意一条包括x节点的简单路径的时候，多增加1就能保持 属性5 。
When we remove or move the black node y, we "push" its blackness onto node x.
当我们移动黑色节点y时，我们将她的黑色推到节点x上。
The problem is that now node x is neither red nor black, thereby violating property 1.
现在问题是现在的节点x既不是黑色也不是红色，违反了属性1.
Instead,node x is either "doubly black" or "red-and-black," and it contributes either 2 or 1, respectively, to the count of black nodes on simple paths containing x.
节点x是双黑，或红黑，在计算包含x的简单路径上它将分别贡献一个或两个。
The color attribute of x will still be either RED (if x is red-and-black) or BLACK (if x is doubly black).
x的属相将仍然是红色（如果x是红黑）或黑色（如果x是双黑）。
In other words, the extra black on a node is reflected in x's pointing to the node rather than in the color attribute.
换句话说，一个节点的额外的黑色反应了x得位置而非颜色属性。
Case1: x'sibling w is red
通过转化，转换成Case2，3，4 的任意一种
Case2 ：x's sibling w is black ,and both of w's children are black
即x的兄弟，还有兄弟的孩子都是黑色，则让x的兄弟便红，x转移到x得父节点
Case3：x的兄弟是黑色，兄弟的右孩子是红色。则通过转换，转换成Case4
Case4：x的兄弟是黑色，兄弟的左孩子是红色。这个就可以解决黑色节点的问题。
