Binary search tree

Binary search tree, also called binary sorting tree, binary search tree, is a binary tree with specific rules, defined as follows:

It is a binary tree, or an empty tree.
The values of all nodes in the left subtree are less than its root node, and the values of all nodes in the right subtree are greater than its root node.
The left and right subtrees are also a binary search tree.

The characteristic of the binary search tree is that looking for the left son down to the left son can find the smallest element, and looking for the right son to the right son can find the largest element.

It seems that we can use it to implement element sorting, but we use binary heap to implement heap sorting, because the binary search tree is not guaranteed to be a balanced binary tree, and the worst case binary search tree will degenerate into A linked list, that is, all nodes have no left subtree or right subtree, the tree hierarchy is too deep, resulting in poor sorting performance.

Using binary search, we can quickly find the value we need in a binary search tree.

Let's analyze the method of adding, deleting, and finding elements in a binary search tree.

First, add elements

The following is a representation of a binary search tree:

// 二叉查找树
type BinarySearchTree struct {
    Root *BinarySearchTreeNode // 树根节点
}

// 二叉查找树节点
type BinarySearchTreeNode struct {
    Value int64                 // 值
    Times int64                 // 值出现的次数
    Left  *BinarySearchTreeNode // 左子树
    Right *BinarySearchTreeNode // 右字树
}

// 初始化一个二叉查找树
func NewBinarySearchTree() *BinarySearchTree {
    return new(BinarySearchTree)
}

A node represents an element, and the Valuevalue of the node is the key used for binary search. When the Valuevalue is repeated, we increase the number of times the value occurs by Times1. The code for adding elements is as follows:

// 添加元素
func (tree *BinarySearchTree) Add(value int64) {
    // 如果没有树根，证明是颗空树，添加树根后返回
    if tree.Root == nil {
        tree.Root = &BinarySearchTreeNode{Value: value}
        return
    }

    // 将值添加进去
    tree.Root.Add(value)
}

func (node *BinarySearchTreeNode) Add(value int64) {
    if value < node.Value {
        // 如果插入的值比节点的值小，那么要插入到该节点的左子树中
        // 如果左子树为空，直接添加
        if node.Left == nil {
            node.Left = &BinarySearchTreeNode{Value: value}
        } else {
            // 否则递归
            node.Left.Add(value)
        }
    } else if value > node.Value {
        // 如果插入的值比节点的值大，那么要插入到该节点的右子树中
        // 如果右子树为空，直接添加
        if node.Right == nil {
            node.Right = &BinarySearchTreeNode{Value: value}
        } else {
            // 否则递归
            node.Right.Add(value)
        }
    } else {
        // 值相同，不需要添加，值出现的次数加1即可
        node.Times = node.Times + 1
    }
}

If it is an empty tree when adding elements, then initialize the root node.

Then the added value is compared with the root node to determine whether it is to be inserted into the left or right subtree of the root node, or not to be inserted.

When the value is smaller than the root node, the element is inserted into the left sub-tree of the root node. When the value is larger than the root node, the element is inserted into the right sub-tree of the root node.

Then recursively operate the left and right subtrees of the root node.

Second, find the maximum or minimum element

Finding the maximum and minimum values is relatively simple. Keep looking for the left son down to the left son to find the smallest element. Keep looking for the right son to the right son to find the largest element.

// 找出最小值的节点
func (tree *BinarySearchTree) FindMinValue() *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.FindMinValue()
}

func (node *BinarySearchTreeNode) FindMinValue() *BinarySearchTreeNode {
    // 左子树为空，表面已经是最左的节点了，该值就是最小值
    if node.Left == nil {
        return node
    }

    // 一直左子树递归
    return node.Left.FindMinValue()
}

// 找出最大值的节点
func (tree *BinarySearchTree) FindMaxValue() *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.FindMaxValue()
}

func (node *BinarySearchTreeNode) FindMaxValue() *BinarySearchTreeNode {
    // 右子树为空，表面已经是最右的节点了，该值就是最大值
    if node.Right == nil {
        return node
    }

    // 一直右子树递归
    return node.Right.FindMaxValue()
}

Third, find the specified element

The binary search technique is also useful here:

// 查找节点
func (tree *BinarySearchTree) Find(value int64) *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.Find(value)
}

func (node *BinarySearchTreeNode) Find(value int64) *BinarySearchTreeNode {
    if value == node.Value {
        // 如果该节点刚刚等于该值，那么返回该节点
        return node
    } else if value < node.Value {
        // 如果查找的值小于节点值，从节点的左子树开始找
        if node.Left == nil {
            // 左子树为空，表示找不到该值了，返回nil
            return nil
        }
        return node.Left.Find(value)
    } else {
        // 如果查找的值大于节点值，从节点的右子树开始找
        if node.Right == nil {
            // 右子树为空，表示找不到该值了，返回nil
            return nil
        }
        return node.Right.Find(value)
    }
}

If it is an empty tree, return nil, otherwise compare with the root node.

If it is just equal to the value of the root node, return the node, otherwise, according to the value comparison, continue to recursively search the left subtree or right word tree.

Fourth, find the father of the specified element

It is the same as searching for the specified element, except that the parent node of the element is tracked.

// 查找指定节点的父亲
func (tree *BinarySearchTree) FindParent(value int64) *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    // 如果根节点等于该值，根节点其没有父节点，返回nil
    if tree.Root.Value == value {
        return nil
    }
    return tree.Root.FindParent(value)
}

func (node *BinarySearchTreeNode) FindParent(value int64) *BinarySearchTreeNode {
    // 外层没有值相等的判定，因为在内层已经判定完毕后返回父亲节点。

    if value < node.Value {
        // 如果查找的值小于节点值，从节点的左子树开始找
        leftTree := node.Left
        if leftTree == nil {
            // 左子树为空，表示找不到该值了，返回nil
            return nil
        }

        // 左子树的根节点的值刚好等于该值，那么父亲就是现在的node，返回
        if leftTree.Value == value {
            return node
        } else {
            return leftTree.FindParent(value)
        }
    } else {
        // 如果查找的值大于节点值，从节点的右子树开始找
        rightTree := node.Right
        if rightTree == nil {
            // 右子树为空，表示找不到该值了，返回nil
            return nil
        }

        // 右子树的根节点的值刚好等于该值，那么父亲就是现在的node，返回
        if rightTree.Value == value {
            return node
        } else {
            return rightTree.FindParent(value)
        }
    }
}

The code has been adjusted accordingly to facilitate obtaining the parent node.

If the returned parent node is empty, it means there is no father.

Five, delete elements

There are four cases of deleting elements:

In the first case, the root node is deleted, and the root node has no son, just delete it directly.
In the second case, the deleted node has a parent node, but there is no sub-tree, that is, the deleted leaf node, just delete it directly.
In the third case, there are two subtrees under the deleted node, because the values of the right subtree are larger than the left subtree, then replace the deleted node with the smallest element in the right subtree, then the nature of the binary search tree Satisfied again. The smallest element of the right subtree can be found as long as you keep looking to the left of the right subtree.
In the fourth case, the deleted node has only one subtree, so the subtree can directly replace the deleted node.

The code is implemented as follows:

// 删除指定的元素
func (tree *BinarySearchTree) Delete(value int64) {
    if tree.Root == nil {
        // 如果是空树，直接返回
        return
    }

    // 查找该值是否存在
    node := tree.Root.Find(value)
    if node == nil {
        // 不存在该值，直接返回
        return
    }

    // 查找该值的父亲节点
    parent := tree.Root.FindParent(value)

    // 第一种情况，删除的是根节点，且根节点没有儿子
    if parent == nil && node.Left == nil && node.Right == nil {
        // 置空后直接返回
        tree.Root = nil
        return
    } else if node.Left == nil && node.Right == nil {
        // 第二种情况，删除的节点有父亲节点，但没有子树

        // 如果删除的是节点是父亲的左儿子，直接将该值删除即可
        if parent.Left != nil && value == parent.Left.Value {
            parent.Left = nil
        } else {
            // 删除的原来是父亲的右儿子，直接将该值删除即可
            parent.Right = nil
        }
        return
    } else if node.Left != nil && node.Right != nil {
        // 第三种情况，删除的节点下有两个子树，因为右子树的值都比左子树大，那么用右子树中的最小元素来替换删除的节点。
        // 右子树的最小元素，只要一直往右子树的左边一直找一直找就可以找到，替换后二叉查找树的性质又满足了。

        // 找右子树中最小的值，一直往右子树的左边找
        minNode := node.Right
        for minNode.Left != nil {
            minNode = minNode.Left
        }
        // 把最小的节点删掉
        tree.Delete(minNode.Value)

        // 最小值的节点替换被删除节点
        node.Value = minNode.Value
        node.Times = minNode.Times
    } else {
        // 第四种情况，只有一个子树，那么该子树直接替换被删除的节点即可

        // 父亲为空，表示删除的是根节点，替换树根
        if parent == nil {
            if node.Left != nil {
                tree.Root = node.Left
            } else {
                tree.Root = node.Right
            }
            return
        }
        // 左子树不为空
        if node.Left != nil {
            // 如果删除的是节点是父亲的左儿子，让删除的节点的左子树接班
            if parent.Left != nil && value == parent.Left.Value {
                parent.Left = node.Left
            } else {
                parent.Right = node.Left
            }
        } else {
            // 如果删除的是节点是父亲的左儿子，让删除的节点的右子树接班
            if parent.Left != nil && value == parent.Left.Value {
                parent.Left = node.Right
            } else {
                parent.Right = node.Right
            }
        }
    }
}

First find the node of the element you want to delete:, tree.Root.Find(value)and then find the father of the node:, tree.Root.FindParent(value)fill the deleted node according to four different situations. The core lies in that in the third case, when the deleted node has two subtrees, it is necessary to replace the deleted node with the smallest node in the right subtree.

The above code can be optimized, you can find out its parent node when searching for the node of the deleted element, it is not necessary to separately query the parent node, in the third case, you can directly remove it when you find the smallest node of the right subtree There is no need to use it recursively tree.Delete(minNode.Value).

Because this general form of binary search tree is rarely used, most programs use AVL trees or red-black trees. The above optimization can be understood.

Six, in-order traversal (to achieve sorting)

Sorting can be achieved using a binary search tree, as long as the tree is traversed in order.

We first print out the left subtree, then print the value of the root node, and then print the right subtree. This is a recursive process.

// 中序遍历
func (tree *BinarySearchTree) MidOrder() {
    tree.Root.MidOrder()
}

func (node *BinarySearchTreeNode) MidOrder() {
    if node == nil {
        return
    }

    // 先打印左子树
    node.Left.MidOrder()

    // 按照次数打印根节点
    for i := 0; i <= int(node.Times); i++ {
        fmt.Println(node.Value)
    }

    // 打印右子树
    node.Right.MidOrder()
}

Seven, complete code

package main

import (
    "fmt"
)

// 二叉查找树节点
type BinarySearchTree struct {
    Root *BinarySearchTreeNode // 树根节点
}

// 二叉查找树节点
type BinarySearchTreeNode struct {
    Value int64                 // 值
    Times int64                 // 值出现的次数
    Left  *BinarySearchTreeNode // 左子树
    Right *BinarySearchTreeNode // 右字树
}

// 初始化一个二叉查找树
func NewBinarySearchTree() *BinarySearchTree {
    return new(BinarySearchTree)
}

// 添加元素
func (tree *BinarySearchTree) Add(value int64) {
    // 如果没有树根，证明是颗空树，添加树根后返回
    if tree.Root == nil {
        tree.Root = &BinarySearchTreeNode{Value: value}
        return
    }

    // 将值添加进去
    tree.Root.Add(value)
}

func (node *BinarySearchTreeNode) Add(value int64) {
    if value < node.Value {
        // 如果插入的值比节点的值小，那么要插入到该节点的左子树中
        // 如果左子树为空，直接添加
        if node.Left == nil {
            node.Left = &BinarySearchTreeNode{Value: value}
        } else {
            // 否则递归
            node.Left.Add(value)
        }
    } else if value > node.Value {
        // 如果插入的值比节点的值大，那么要插入到该节点的右子树中
        // 如果右子树为空，直接添加
        if node.Right == nil {
            node.Right = &BinarySearchTreeNode{Value: value}
        } else {
            // 否则递归
            node.Right.Add(value)
        }
    } else {
        // 值相同，不需要添加，值出现的次数加1即可
        node.Times = node.Times + 1
    }
}

// 找出最小值的节点
func (tree *BinarySearchTree) FindMinValue() *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.FindMinValue()
}

func (node *BinarySearchTreeNode) FindMinValue() *BinarySearchTreeNode {
    // 左子树为空，表面已经是最左的节点了，该值就是最小值
    if node.Left == nil {
        return node
    }

    // 一直左子树递归
    return node.Left.FindMinValue()
}

// 找出最大值的节点
func (tree *BinarySearchTree) FindMaxValue() *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.FindMaxValue()
}

func (node *BinarySearchTreeNode) FindMaxValue() *BinarySearchTreeNode {
    // 右子树为空，表面已经是最右的节点了，该值就是最大值
    if node.Right == nil {
        return node
    }

    // 一直右子树递归
    return node.Right.FindMaxValue()
}

// 查找指定节点
func (tree *BinarySearchTree) Find(value int64) *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    return tree.Root.Find(value)
}

func (node *BinarySearchTreeNode) Find(value int64) *BinarySearchTreeNode {
    if value == node.Value {
        // 如果该节点刚刚等于该值，那么返回该节点
        return node
    } else if value < node.Value {
        // 如果查找的值小于节点值，从节点的左子树开始找
        if node.Left == nil {
            // 左子树为空，表示找不到该值了，返回nil
            return nil
        }
        return node.Left.Find(value)
    } else {
        // 如果查找的值大于节点值，从节点的右子树开始找
        if node.Right == nil {
            // 右子树为空，表示找不到该值了，返回nil
            return nil
        }
        return node.Right.Find(value)
    }
}

// 查找指定节点的父亲
func (tree *BinarySearchTree) FindParent(value int64) *BinarySearchTreeNode {
    if tree.Root == nil {
        // 如果是空树，返回空
        return nil
    }

    // 如果根节点等于该值，根节点其没有父节点，返回nil
    if tree.Root.Value == value {
        return nil
    }
    return tree.Root.FindParent(value)
}

func (node *BinarySearchTreeNode) FindParent(value int64) *BinarySearchTreeNode {
    // 外层没有值相等的判定，因为在内层已经判定完毕后返回父亲节点。

    if value < node.Value {
        // 如果查找的值小于节点值，从节点的左子树开始找
        leftTree := node.Left
        if leftTree == nil {
            // 左子树为空，表示找不到该值了，返回nil
            return nil
        }

        // 左子树的根节点的值刚好等于该值，那么父亲就是现在的node，返回
        if leftTree.Value == value {
            return node
        } else {
            return leftTree.FindParent(value)
        }
    } else {
        // 如果查找的值大于节点值，从节点的右子树开始找
        rightTree := node.Right
        if rightTree == nil {
            // 右子树为空，表示找不到该值了，返回nil
            return nil
        }

        // 右子树的根节点的值刚好等于该值，那么父亲就是现在的node，返回
        if rightTree.Value == value {
            return node
        } else {
            return rightTree.FindParent(value)
        }
    }
}

// 删除指定的元素
func (tree *BinarySearchTree) Delete(value int64) {
    if tree.Root == nil {
        // 如果是空树，直接返回
        return
    }

    // 查找该值是否存在
    node := tree.Root.Find(value)
    if node == nil {
        // 不存在该值，直接返回
        return
    }

    // 查找该值的父亲节点
    parent := tree.Root.FindParent(value)

    // 第一种情况，删除的是根节点，且根节点没有儿子
    if parent == nil && node.Left == nil && node.Right == nil {
        // 置空后直接返回
        tree.Root = nil
        return
    } else if node.Left == nil && node.Right == nil {
        //  第二种情况，删除的节点有父亲节点，但没有子树

        // 如果删除的是节点是父亲的左儿子，直接将该值删除即可
        if parent.Left != nil && value == parent.Left.Value {
            parent.Left = nil
        } else {
            // 删除的原来是父亲的右儿子，直接将该值删除即可
            parent.Right = nil
        }
        return
    } else if node.Left != nil && node.Right != nil {
        // 第三种情况，删除的节点下有两个子树，因为右子树的值都比左子树大，那么用右子树中的最小元素来替换删除的节点，这时二叉查找树的性质又满足了。

        // 找右子树中最小的值，一直往右子树的左边找
        minNode := node.Right
        for minNode.Left != nil {
            minNode = minNode.Left
        }
        // 把最小的节点删掉
        tree.Delete(minNode.Value)

        // 最小值的节点替换被删除节点
        node.Value = minNode.Value
        node.Times = minNode.Times
    } else {
        // 第四种情况，只有一个子树，那么该子树直接替换被删除的节点即可

        // 父亲为空，表示删除的是根节点，替换树根
        if parent == nil {
            if node.Left != nil {
                tree.Root = node.Left
            } else {
                tree.Root = node.Right
            }
            return
        }
        // 左子树不为空
        if node.Left != nil {
            // 如果删除的是节点是父亲的左儿子，让删除的节点的左子树接班
            if parent.Left != nil && value == parent.Left.Value {
                parent.Left = node.Left
            } else {
                parent.Right = node.Left
            }
        } else {
            // 如果删除的是节点是父亲的左儿子，让删除的节点的右子树接班
            if parent.Left != nil && value == parent.Left.Value {
                parent.Left = node.Right
            } else {
                parent.Right = node.Right
            }
        }
    }
}

// 中序遍历
func (tree *BinarySearchTree) MidOrder() {
    tree.Root.MidOrder()
}

func (node *BinarySearchTreeNode) MidOrder() {
    if node == nil {
        return
    }

    // 先打印左子树
    node.Left.MidOrder()

    // 按照次数打印根节点
    for i := 0; i <= int(node.Times); i++ {
        fmt.Println(node.Value)
    }

    // 打印右子树
    node.Right.MidOrder()
}

func main() {
    values := []int64{3, 6, 8, 20, 9, 2, 6, 8, 9, 3, 5, 40, 7, 9, 13, 6, 8}

    // 初始化二叉查找树并添加元素
    tree := NewBinarySearchTree()
    for _, v := range values {
        tree.Add(v)
    }

    // 找到最大值或最小值的节点
    fmt.Println("find min value:", tree.FindMinValue())
    fmt.Println("find max value:", tree.FindMaxValue())

    // 查找不存在的99
    node := tree.Find(99)
    if node != nil {
        fmt.Println("find it 99!")
    } else {
        fmt.Println("not find it 99!")
    }

    // 查找存在的9
    node = tree.Find(9)
    if node != nil {
        fmt.Println("find it 9!")
    } else {
        fmt.Println("not find it 9!")
    }

    // 删除存在的9后，再查找9
    tree.Delete(9)
    node = tree.Find(9)
    if node != nil {
        fmt.Println("find it 9!")
    } else {
        fmt.Println("not find it 9!")
    }

    // 中序遍历，实现排序
    tree.MidOrder()
}

After running the program, the result:

find min value: &{2 0 <nil> <nil>}
find max value: &{40 0 <nil> <nil>}
not find it 99!
find it 9!
not find it 9!
2
3
3
5
6
6
6
7
8
8
8
13
20
40

8. Summary

The binary search tree may degenerate into a linked list, or it may be a very balanced binary tree. The time complexity of searching, adding, and deleting elements depends on the height of the tree h.

When the binary tree is full, the height of the tree is the smallest, at this time the number of tree nodes nand height hrelationship is as follows: h = log(n).
When a binary tree list, the number of tree nodes at this time nand the height hrelationship is: h = n.

The efficiency of the binary search tree stems from the characteristics of its binary search. The time complexity lies in the height of the binary tree, so the time complexity range of searching, adding and deleting is log(n)~n.

In order to improve the search speed of the binary search tree, the height of the tree should be as small as possible. The AVL tree and the red-black tree are relatively balanced binary search trees, because of the special rotation balancing operation, the height of the tree is greatly reduced. They have high search efficiency, and the average time complexity of add, delete, and search operations are all log(n), and they are often used in various programs.

The binary search tree is the basis of the advanced data structure AVL tree, red-black tree to be learned later.

Series article entry

I am the star Chen, Welcome I have personally written data structures and algorithms (Golang achieve) , starting in the article to read more friendly GitBook .

Data structure and algorithm (Golang implementation) (27) Search algorithm-binary search tree