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@@ -9,15 +9,15 @@ As shown in the figure below, a <u>binary search tree</u> satisfies the followin
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## Operations on a binary search tree
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We encapsulate the binary search tree as a class `BinarySearchTree` and declare a member variable `root`, pointing to the tree's root node.
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We encapsulate the binary search tree as a class `BinarySearchTree` and declare a member variable `root` pointing to the tree's root node.
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### Searching for a node
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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` and start from the binary tree's root node `root`, looping to compare the size relationship between the node value `cur.val` and `num`.
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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`.
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- If `cur.val < num`, it means the target node is in `cur`'s right subtree, thus execute `cur = cur.right`.
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- If `cur.val > num`, it means the target node is in `cur`'s left subtree, thus execute `cur = cur.left`.
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- If `cur.val = num`, it means the target node is found, exit the loop and return the node.
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- If `cur.val = num`, it means the target node is found, exit the loop, and return the node.
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=== "<1>"
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The search operation in a binary search tree works on the same principle as the binary search algorithm, eliminating half of the possibilities 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. Example code is as follows:
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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:
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Given an element `num` to be inserted, to maintain the property of the binary search tree "left subtree < root node < right subtree," the insertion operation proceeds as shown in the figure below.
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1.**Finding the insertion position**: Similar to the search operation, start from the root node and loop downwards according to the size relationship between the current node value and `num` until passing through the leaf node (traversing to `None`) then exit the loop.
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2.**Insert the node at that position**: Initialize the node `num` and place it where `None` was.
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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.
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2.**Insert the node at this position**: Initialize the node `num` and place it where `None` was.
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In the code implementation, note the following two points.
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- The binary search tree does not allow duplicate nodes; otherwise, it will violate its definition. Therefore, if the node to be inserted already exists in the tree, the insertion is not performed, and it directly returns.
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- To perform the insertion operation, we need to use the node `pre` to save the node from the last loop. This way, when traversing to `None`, we can get its parent node, thus completing the node insertion operation.
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- 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.
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- 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.
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### Removing a node
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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 < root node < right subtree" is still satisfied. Therefore, based on the number of child nodes of the target node, we divide it into 0, 1, and 2 cases, performing the corresponding node removal operations.
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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 < root node < right subtree" is still satisfied. Therefore, based on the number of child nodes of the target node, we divide it into three cases: 0, 1, and 2, and perform the corresponding node removal operations.
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As shown in the figure below, when the degree of the node to be removed is $0$, it means the node is a leaf node, and it can be directly removed.
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As shown in the figure below, when the degree of the node to be removed is $0$, it means the node is a leaf node and can be directly removed.
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### In-order traversal is ordered
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As shown in the figure below, the in-order traversal of a binary tree follows the "left $\rightarrow$ root $\rightarrow$ right" traversal order, and a binary search tree satisfies the size relationship "left child node $<$ root node $<$ right child node".
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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."
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This means that in-order traversal in a binary search tree always traverses the next smallest node first, thus deriving an important property: **The in-order traversal sequence of a binary search tree is ascending**.
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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**.
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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.
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## Efficiency of binary search trees
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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. Only in scenarios of high-frequency addition and low-frequency search and removal, arrays are more efficient than binary search trees.
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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.
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<palign="center"> Table <id> Efficiency comparison between arrays and search trees </p>
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| Insert element | $O(1)$ | $O(\log n)$ |
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| Remove element | $O(n)$ | $O(\log n)$ |
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In ideal conditions, the binary search tree is "balanced," thus any node can be found within $\log n$ loops.
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Ideally, the binary search tree is "balanced," allowing any node can be found within $\log n$ loops.
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However, continuously inserting and removing nodes in a binary search treemay lead to the binary tree degenerating into a chain list as shown in the figure below, at which point the time complexity of various operations also degrades to $O(n)$.
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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 $O(n)$.
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