bst 刪除節(jié)點

Problem statement:

問題陳述：

Given a BST and a value x, write a function to delete the nodes having values greater than or equal to x. The function will return the modified root.

給定一個BST和一個值x ，編寫一個函數(shù)刪除值大于或等于x的節(jié)點 。 該函數(shù)將返回修改后的根。

Example:

例：

Input BST8/ \4 10/ \ / \3 5 9 11Input X: 10After deletion:8/ \4 9/ \ 3 5

Solution:

解：

Basic properties of BST:

BST的基本屬性：

All node values are distinct.

所有節(jié)點值都是不同的。

The left child node is always smaller than the root & right child node is always greater than the root.

左子節(jié)點始終小于根節(jié)點，右子節(jié)點始終大于根節(jié)點。

Thus it's clear whenever a root has a value greater than or equal to the input value X, the entire right subtree and root need to be deleted, but not the left one.

因此，很明顯，只要根的值大于或等于輸入值X ，就需要刪除整個右子樹和根，而不必刪除左子樹和根。

Algorithm:

算法：

FUNCTION buildTree (root,X)IF (root==NULL)Return NULL;END IFIF (root->data is equal to X)return root->left; //returning the left subtree basicallyELSE IF(root->data>key)//recur for the left subtree since left subtree may also have nodes //that need to be deleted due to greater or equal valuesreturn buildTree(root->left, X); ELSE//root is not at all to be deleted, same for left subtree, //recursively process right subtreeroot->right=buildTree(root->right, X);return root; END FUNCTION

The above function actually builds the tree deleting the nodes having greater and equal value than X.

上面的函數(shù)實際上是在構(gòu)建樹，刪除值大于和等于X的節(jié)點。

Example with Explanation:

解釋示例：

Nodes are represented by respected values for better understanding8/ \4 10/ \ / \3 5 9 11Input X: 10At main we callbuildTree (8, 10) ------------------------------------------------buildTree (8, 10) root not NULL 8<10 Thus root->left= 8->left root->right=buildTree (8->right, 10) //buildTree (10, 10) ------------------------------------------------buildTree (10, 10) root not NULL 10==10 Thus return root->left= 10->left=9 ------------------------------------------------Hence At buildTree (8, 10) Thus root->left= 8->left root->right=9 (9 is the root to the remaining subtree which is NULL here luckily)So the tree actually becomes8/ \4 9/ \ 3 5 .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

C++ Implementation:

C ++實現(xiàn)：

#include <bits/stdc++.h> using namespace std;class Node{public: int data; //valueNode *left; //pointer to left childNode *right; //pointer to right child };// creating new node Node* newnode(int data) { Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = NULL; node->right = NULL; return(node); }Node* buildTree(Node* root,int key){if(!root)return NULL;if(root->data==key){return root->left; //only left subtree not deleted}else if(root->data>key)return buildTree(root->left,key); //recur for left subtreeelse{//root not deleted//left subtree not deleted//buld right subtreeroot->right=buildTree(root->right,key); return root;} }Node* deleteNode(Node* root, int key) {root=buildTree(root,key);return root; }void levelwisedisplay(Node *root) {//to display the tree level wisequeue<Node*> q;Node* temp;int count=0;q.push(root);q.push(NULL);while(!q.empty()){temp=q.front();q.pop();if(temp==NULL){if(!q.empty())q.push(NULL);cout<<"\nend of level "<<count++<<endl;}else{cout<<temp->data<<" ";if(temp->left)q.push(temp->left);if(temp->right)q.push(temp->right);}}return ; }int main(){cout<<"tree is built as per example\n";Node *root=newnode(8); root->left= newnode(4); root->right= newnode(10); root->right->right=newnode(11);root->right->left=newnode(9);root->left->left=newnode(3); root->left->right=newnode(5);printf("Displaying level wise\n");levelwisedisplay(root);root=deleteNode(root,10);//as per exampleprintf("after deleting nodes greater or equal to 10\n");levelwisedisplay(root);return 0; }

Output

輸出量

tree is built as per example Displaying level wise 8 end of level 0 4 10 end of level 1 3 5 9 11 end of level 2 after deleting nodes greater or equal to 10 8 end of level 0 4 9 end of level 1 3 5 end of level 2

翻譯自: https://www.includehelp.com/icp/delete-nodes-greater-than-or-equal-to-k-in-a-bst.aspx

bst 刪除節(jié)點

總結(jié)

以上是生活随笔為你收集整理的bst 删除节点_在BST中删除大于或等于k的节点的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。