सबट्री हिंदी उपयोग और उदाहरण

इंसर्शन शुरू होती है जैसे ही खोज शुरू होती हैं यदि कुंजी जड़ के बराबर नहीं है, तो हम पहले की तरह बाएं या दाएं सबट्री खोजते हैं।

इसे पहले रूट के पूरे बाएँ उप-भाग के माध्यम से पुनरावृत्ति करते हुए और फिर पुन: रूट के माध्यम से दाईं सबट्री पर जाकर किया जा सकता है।

""यदि कुंजी का मान रूट नोड के मान से कम है, तो हम बाईं सबट्री में पुनरावर्ती खोज करते हैं और यदि यह इसके मूल्य से अधिक है, तो हम दाए उपप्रकार में खोज करते हैं।

सबट्री इसके अंग्रेजी अर्थ का उदाहरण

All strings in the middle subtree of a node start with that prefix.

At each step the two closest clusters i and j are merged, this procedure repeats until only one cluster remains, each subtree is stored as a subset \tau\in T_\text{LT}.

In the following figures, simple circle nodes represent vertices, double circle nodes represent subtrees, twisted segments represent tree paths, and curved arcs represent cotree edges.

Indeed, this follows from the fact that G has an invariant subtree on which it acts with compact quotient, and the Svarc—Milnor lemma.

, valid for all feasible integer solutions, or local cuts, meaning that they are satisfied by all solutions fulfilling the side constraints from the currently considered branch and bound subtree.

The following binary search tree (BST) corresponds to each execution of quicksort: the initial pivot is the root node; the pivot of the left half is the root of the left subtree, the pivot of the right half is the root of the right subtree, and so on.

meld: compare the two root elements, the smaller remains the root of the result, the larger element and its subtree is appended as a child of this root.

decrease-key (optional): remove the subtree rooted at the key to be decreased, replace the key with a smaller key, then meld the result back into the heap.

delete-min: remove the root and do repeated melds of its subtrees until one tree remains.

The root node may have zero or more subtrees.

The k-th subtree is recursively built of all elements b such that d(a,b) k.

Let the root of r be the root of (the smaller root), and let r's right subtree be p's left subtree.

Now, compute r's left subtree by recursively merging p's right subtree with q.