我知道有很多方法可以构建从前序遍历树（作为一个数组）。 比较常见的问题是构建它，因为中序和预购遍历。 在这种情况下，尽管序遍历是多余的，它绝对会让事情变得更容易。 任何人都可以给我一个想法如何做到这一点的后序遍历？ 重复的和递归的解决方案是必需的。
我试图做到这一点反复进行利用堆栈，但不能在所有获得逻辑正确，所以得到了一个可怕的混乱的树。 同去的递归。
Answer 1:
如果你有从BST的后序遍历数组，你知道这根是数组的最后一个元素。 根的左子占据了阵列的第一部分，和由比所述根较小的条目。 然后如下右子，其由比所述根较大的元件。 （两个孩子可能是空的）。
________________________________
| | |R|
--------------------------------
left child right child root
所以，主要的问题是要找到点左子结束，权开始。
两个孩子也从他们的后序遍历获得的，所以构造它们被以同样的方式完成的，递归。
BST fromPostOrder(value[] nodes) {
// No nodes, no tree
if (nodes == null) return null;
return recursiveFromPostOrder(nodes, 0, nodes.length - 1);
}
// Construct a BST from a segment of the nodes array
// That segment is assumed to be the post-order traversal of some subtree
private BST recursiveFromPostOrder(value[] nodes,
int leftIndex, int rightIndex) {
// Empty segment -> empty tree
if (rightIndex < leftIndex) return null;
// single node -> single element tree
if (rightIndex == leftIndex) return new BST(nodes[leftIndex]);
// It's a post-order traversal, so the root of the tree
// is in the last position
value rootval = nodes[rightIndex];
// Construct the root node, the left and right subtrees are then
// constructed in recursive calls, after finding their extent
BST root = new BST(rootval);
// It's supposed to be the post-order traversal of a BST, so
// * left child comes first
// * all values in the left child are smaller than the root value
// * all values in the right child are larger than the root value
// Hence we find the last index in the range [leftIndex .. rightIndex-1]
// that holds a value smaller than rootval
int leftLast = findLastSmaller(nodes, leftIndex, rightIndex-1, rootval);
// The left child occupies the segment [leftIndex .. leftLast]
// (may be empty) and that segment is the post-order traversal of it
root.left = recursiveFromPostOrder(nodes, leftIndex, leftLast);
// The right child occupies the segment [leftLast+1 .. rightIndex-1]
// (may be empty) and that segment is the post-order traversal of it
root.right = recursiveFromPostOrder(nodes, leftLast + 1, rightIndex-1);
// Both children constructed and linked to the root, done.
return root;
}
// find the last index of a value smaller than cut in a segment of the array
// using binary search
// supposes that the segment contains the concatenation of the post-order
// traversals of the left and right subtrees of a node with value cut,
// in particular, that the first (possibly empty) part of the segment contains
// only values < cut, and the second (possibly empty) part only values > cut
private int findLastSmaller(value[] nodes, int first, int last, value cut) {
// If the segment is empty, or the first value is larger than cut,
// by the assumptions, there is no value smaller than cut in the segment,
// return the position one before the start of the segment
if (last < first || nodes[first] > cut) return first - 1;
int low = first, high = last, mid;
// binary search for the last index of a value < cut
// invariants: nodes[low] < cut
// (since cut is the root value and a BST has no dupes)
// and nodes[high] > cut, or (nodes[high] < cut < nodes[high+1]), or
// nodes[high] < cut and high == last, the latter two cases mean that
// high is the last index in the segment holding a value < cut
while (low < high && nodes[high] > cut) {
// check the middle of the segment
// In the case high == low+1 and nodes[low] < cut < nodes[high]
// we'd make no progress if we chose mid = (low+high)/2, since that
// would then be mid = low, so we round the index up instead of down
mid = low + (high-low+1)/2;
// The choice of mid guarantees low < mid <= high, so whichever
// case applies, we will either set low to a strictly greater index
// or high to a strictly smaller one, hence we won't become stuck.
if (nodes[mid] > cut) {
// The last index of a value < cut is in the first half
// of the range under consideration, so reduce the upper
// limit of that. Since we excluded mid as a possible
// last index, the upper limit becomes mid-1
high = mid-1;
} else {
// nodes[mid] < cut, so the last index with a value < cut is
// in the range [mid .. high]
low = mid;
}
}
// now either low == high or nodes[high] < cut and high is the result
// in either case by the loop invariants
return high;
}
Answer 2:
你并不真正需要的序遍历。 有一个简单的方法来重建只给出了后序遍历的树：
- 取所述输入阵列中的最后一个元素。 这是根。
- 环在剩余的输入阵列寻找其中元素从比所述根小，以成为更大的改变的点。 拆分在该点的输入数组。 这也可以用二进制搜索算法来完成。
- 递归重构来自这两个子阵列的子树。
这可以很容易要么递归或迭代完成了一个堆栈，你可以用两个指标来表示当前的子阵列的开始和结束，而不是实际分裂阵列。
Answer 3:
后序遍历是这样的：
visit left
visit right
print current.
而序是这样的：
visit left
print current
visit right
让我们举个例子：
7
/ \
3 10
/ \ / \
2 5 9 12
/
11
序是： 2 3 5 7 9 10 11 12
邮购是： 2 5 3 9 11 12 10 7
迭代以相反的顺序后序阵列和保持分裂周围，其中该值是中序阵列。 递归地做到这一点，那将是你的树。 例如：
current = 7, split inorder at 7: 2 3 5 | 9 10 11 12
看起来熟悉？ 什么是左边是左子树，什么是右侧是右子树，以伪随机顺序尽可能的BST结构而言。 但是，你现在知道你的根是什么。 现在做了两半相同。 找到一个元件从左半在后序遍历的第一次出现（从端部）。 这将是3约为3斯普利特：
current = 3, split inorder at 3: 2 | 5 ...
所以，你知道你的树是这个样子至今：
7
/
3
这是基于，在后序遍历的值总是会出现在其子女出现后，在序遍历的值将其子值之间出现的事实。
Answer 4:
不要遍历任何东西。 最后一个元素是你的根。 然后取阵列向后遵循BST的插入规则。
eg:-
given just the postorder -- 2 5 3 9 11 12 10 7
7
\
10
----
7
\
10
\
12
-----
7
\
10
\
12
/
11
-------
7
\
10
/ \
9 12
/
11
--------
7
/ \
3 10
/ \ / \
2 5 9 12
/
11
文章来源: How to construct BST given post-order traversal
