Given a target node value `num`, one can search according to the properties of the binary search tree. As shown in the figure below, we declare a node `cur`, start from the binary tree's root node `root`, and loop to compare the size between the node value `cur.val` and `num`.
The search operation in a binary search tree works on the same principle as the binary search algorithm, eliminating half of the cases in each round. The number of loops is at most the height of the binary tree. When the binary tree is balanced, it uses $O(\log n)$ time. The example code is as follows:
Given an element `num` to be inserted, to maintain the property of the binary search tree "left subtree <rootnode<rightsubtree,"theinsertionoperationproceedsasshowninthefigurebelow.
1.**Finding insertion position**: Similar to the search operation, start from the root node, loop downwards according to the size relationship between the current node value and `num`, until the leaf node is passed (traversed to `None`), then exit the loop.
2.**Insert the node at this position**: Initialize the node `num` and place it where `None` was.
- The binary search tree does not allow duplicate nodes to exist; otherwise, its definition would be violated. Therefore, if the node to be inserted already exists in the tree, the insertion is not performed, and the node returns directly.
- To perform the insertion operation, we need to use the node `pre` to save the node from the previous loop. This way, when traversing to `None`, we can get its parent node, thus completing the node insertion operation.
First, find the target node in the binary tree, then remove it. Similar to inserting a node, we need to ensure that after the removal operation is completed, the property of the binary search tree "left subtree <rootnode<rightsubtree"isstillsatisfied.Therefore,basedonthenumberofchildnodesofthetargetnode,wedivideitintothreecases:0,1,and2,andperformthecorrespondingnoderemovaloperations.
When the degree of the node to be removed is $2$, we cannot remove it directly, but need to use a node to replace it. To maintain the property of the binary search tree "left subtree $<$ root node $<$ right subtree," **this node can be either the smallest node of the right subtree or the largest node of the left subtree**.
Assuming we choose the smallest node of the right subtree (the next node in in-order traversal), then the removal operation proceeds as shown in the figure below.
1. Find the next node in the "in-order traversal sequence" of the node to be removed, denoted as `tmp`.
2. Replace the value of the node to be removed with `tmp`'s value, and recursively remove the node `tmp` in the tree.
=== "<1>"
The operation of removing a node also uses $O(\log n)$ time, where finding the node to be removed requires $O(\log n)$ time, and obtaining the in-order traversal successor node requires $O(\log n)$ time. Example code is as follows:
As shown in the figure below, the in-order traversal of a binary tree follows the traversal order of "left $\rightarrow$ root $\rightarrow$ right," and a binary search tree satisfies the size relationship of "left child node $<$ root node $<$ right child node."
This means that when performing in-order traversal in a binary search tree, the next smallest node will always be traversed first, thus leading to an important property: **The sequence of in-order traversal in a binary search tree is ascending**.
Using the ascending property of in-order traversal, obtaining ordered data in a binary search tree requires only $O(n)$ time, without the need for additional sorting operations, which is very efficient.
Given a set of data, we consider using an array or a binary search tree for storage. Observing the table below, the operations on a binary search tree all have logarithmic time complexity, which is stable and efficient. Arrays are more efficient than binary search trees only in scenarios involving frequent additions and infrequent searches or removals.
However, if we continuously insert and remove nodes in a binary search tree, it may degenerate into a linked list as shown in the figure below, where the time complexity of various operations also degrades to $O(n)$.
