我知道有一些方法可以从预序遍历(作为数组)构建树。更常见的问题是在给定中序和先序遍历的情况下构造它。在这种情况下,中序遍历虽然是多余的,但绝对让事情变得简单。谁能告诉我如何进行后序遍历?需要迭代和递归解决方案。
我尝试使用堆栈迭代地完成它,但根本无法正确理解逻辑,所以得到了一棵可怕的凌乱树。递归也是如此。
最佳答案
如果您有一个来自 BST 后序遍历的数组,您知道根是数组的最后一个元素。根的左 child 占据数组的第一部分,并且由小于根的条目组成。然后是右 child ,由大于根的元素组成。 (两个 child 都可能是空的)。
________________________________
| | |R|
--------------------------------
left child right child root
所以主要的问题是找到左 child 结束和右 child 开始的点。
这两个 child 也是从他们的后序遍历中获得的,所以构造他们也是以同样的方式递归完成的。
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;
}
关于algorithm - 给定后序遍历如何构建BST,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13167536/