转换二叉树preorder上市后序，反之亦然 [英] Convert Preorder listing of a binary tree to postorder and vice versa
问题描述
我如何才能找到一棵树的preorder上市如果只是后序上市，并给出反之亦然。此外，在树中，每个非叶结点有两个孩子（即每个节点有两个或零儿童。）
How can I find the preorder listing of a tree if only the postorder listing is given and vice versa. Also, in the tree, every non-leaf node has two children (i.e. Each node has either two or zero children.)
编辑：另一个给定的假设是，各节点的标签是唯一的并具有将其标识为内部节点或叶的字段。我认为，应该摆脱单一preorder的模糊性或后序能够唯一标识一棵树。
Another given assumption is that the label of each node is unique and has a field that will identify it as an internal node or a leaf. I think that ought to get rid of the ambiguity of a single preorder or postorder being able to uniquely identify a tree.
推荐答案
在不假设树的节点有idenrify自己内部或叶的字段，你找不到你的问题了独特的答案。这种假设或序房源必须是可用找到一种独特的树。 在这种情况下，找到一个可能的答案，你可以构造如下图所示，以匹配任何给定的后序上市形式的树：（假设后序列表：1 2 3 4 5 6 7 8 9）
Without assuming that nodes in the tree have a field that idenrify themselves as internal or leaf, you can't find a unique answer for your question. That assumption or inorder listing must be available to find a unique tree. In this case, to find one possible answer, you can construct a tree of the form shown below to match any given postorder listing: (suppose postorder listing is: 1 2 3 4 5 6 7 8 9)
9[7[5[3[1,2],4],6],8]
现在你可以让preorder使用这个树状列表。
Now you can make preorder listing using this tree.
假设树的节点有自己的身份为内部或叶一个领域，我们可以做一个独特的树了后序上市对于这种用这种算法树：
Assuming nodes in the tree have a field that identify themselves as internal or leaf, we can make a unique tree out of a postorder listing for this kind of trees with this algorithm:
- 扫描从后序列表的开头和找到的第一个内部节点。该节点将有整整两个叶片孩子其中precede在后序上市这一节点。
- 在树形结构添加内部节点，并提出两点preceding节点上市的孩子。
- 从列表中删除这两个孩子，使该内部节点叶。
- 转到步骤1，重复直到上市变为空。
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