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Warning: -file- is being assigned a //# sourceMappingURL, but already has one.To construct the complete binary search tree, recursively set the range for each recursive call and return if the next element of the postorder traversal is out of the valid range. Suppose the root node has value x, recur for the right subtree with range (x, INFINITY) and recur for the left subtree with range [-INFINITY, x). Like the previous approach, construct BST’s root node from the last item in the postorder sequence. It means that the root node and any of its children can have keys ranging between -INFINITY and INFINITY. We start by setting the range as for the root node. The idea to pass the information regarding the valid range of keys for the current root node and its children in the recursion itself. We know that each node has a key that is greater than all keys present in its left subtree and less than the keys present in the right subtree in the BST. We can reduce the time complexity to O(n) by following a different approach that doesn’t involve searching for an index that separates the left and right subtree keys in a postorder sequence. The time complexity of the above solution is O(n 2), where n is the size of the BST, and requires space proportional to the tree’s height for the call stack. Inorder traversal of BST is 8 10 12 15 16 20 25 The algorithm can be implemented as follows in C, Java, and Python: Given a distinct sequence of keys representing the postorder traversal of a binary search tree, construct a BST from it.įor example, the following BST should be constructed for postorder traversal.
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