本篇內(nèi)容介紹了“C++中如何實(shí)現(xiàn)搜索二叉樹”的有關(guān)知識(shí),在實(shí)際案例的操作過(guò)程中,不少人都會(huì)遇到這樣的困境,接下來(lái)就讓小編帶領(lǐng)大家學(xué)習(xí)一下如何處理這些情況吧!希望大家仔細(xì)閱讀,能夠?qū)W有所成!
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二叉查找樹(英語(yǔ):Binary Search Tree),也稱二叉搜索樹、有序二叉樹(英語(yǔ):ordered binary tree),排序二叉樹(英語(yǔ):sorted binary tree),是指一棵空樹或者具有下列性質(zhì)的二叉樹:
任意節(jié)點(diǎn)的左子樹不空,則左子樹上所有結(jié)點(diǎn)的值均小于它的根結(jié)點(diǎn)的值;
任意節(jié)點(diǎn)的右子樹不空,則右子樹上所有結(jié)點(diǎn)的值均大于它的根結(jié)點(diǎn)的值;
任意節(jié)點(diǎn)的左、右子樹也分別為二叉查找樹;
沒(méi)有鍵值相等的節(jié)點(diǎn)。
#pragma once templatestruct BSTreeNode { K _key; V _value; BSTreeNode * _left; BSTreeNode * _right; BSTreeNode(const K& key, const V& value) :_key(key) ,_value(value) ,_left(NULL) ,_right(NULL) {} }; template class BSTree { typedef BSTreeNode Node; public: BSTree() :_root(NULL) {} bool Insert(const K& key, const V& value) { if (NULL == _root)//若為空樹 { _root = new Node(key, value); return true; } Node* parent = NULL; Node* cur = _root; //確定插入節(jié)點(diǎn)的位置 while (cur) { if (key < cur->_key) { parent = cur; cur = cur->_left; } else if (key > cur->_key) { parent = cur; cur = cur->_right; } else//已經(jīng)存在key { return false; } } //插入節(jié)點(diǎn) if (key > parent->_key) parent->_right = new Node(key, value); else parent->_left = new Node(key, value); } //Insert遞歸寫法 bool InsertR(const K& key, const V& value) { return _InsertR(_root, key, value); } bool _InsertR(Node*& root, const K& key, const V& value) { if (NULL == root) { root = new Node(key, value); return true; } if (key > root->_key) return _InsertR(root->_right, key, value); else if (key < root->_key) return _InsertR(root->_left, key, value); else return false; } Node* Find(const K& key) { Node* cur = _root; while (cur) { if (key > cur->_key) cur = cur->_right; else if (key < cur->_key) cur = cur->_left; else return cur; } return NULL; } //Find遞歸寫法 Node* FindR(const K& key) { return _FindR(_root, key); } Node* _FindR(Node* root, const K& key) { if (NULL == root) return NULL; if (key > root->_key) return _FindR(root->_right, key); else if (key < root->_key) return _FindR(root->_left, key); else return root; } bool Remove(const K& key) { Node* parent = NULL; Node* cur = _root; //確定刪除節(jié)點(diǎn)的位置 while (cur) { if (key > cur->_key) { parent = cur; cur = cur->_right; } else if (key < cur->_key) { parent = cur; cur = cur->_left; } else { break; } } if (NULL == cur)//沒(méi)有該節(jié)點(diǎn) { return false; } Node* del; if (NULL == cur->_left)//刪除節(jié)點(diǎn)的左孩子為空 { del = cur; //刪除的節(jié)點(diǎn)為根節(jié)點(diǎn) if (NULL == parent) { _root = _root->_right; } else { if (cur == parent->_left) parent->_left = cur->_right; else parent->_right = cur->_right; } } else if (NULL == cur->_right)//刪除節(jié)點(diǎn)的右孩子為空 { del = cur; if (NULL == parent) { _root = _root->_left; } else { if (cur == parent->_left) parent->_left = cur->_right; else parent->_right = cur->_left; } } else//刪除節(jié)點(diǎn)的左右孩子都不為空,找右子樹最左節(jié)點(diǎn)代替該節(jié)點(diǎn)刪除 { parent = cur; Node* leftmost = cur->_right; while (leftmost->_left) { parent = leftmost; leftmost = leftmost->_left; } del = leftmost; cur->_key = leftmost->_key; cur->_value = leftmost->_value; if (leftmost == parent->_left) parent->_left = leftmost->_right; else parent->_right = leftmost->_right; } return true; } //Remove遞歸寫法 bool RemoveR(const K& key) { return _RemoveR(_root, key); } bool _RemoveR(Node*& root, const K& key) { if (NULL == root) return false; if (key > root->_key) { return _RemoveR(root->_right, key); } else if (key < root->_key) { return _RemoveR(root->_left, key); } else { Node* del = root; if (NULL == root->_left) { root = root->_right; } else if (NULL == root->_right) { root = root->_left; } else { Node* leftmost = root->_right; while (leftmost->_left) { leftmost = leftmost->_left; } swap(root->_key, leftmost->_key); swap(root->_value, leftmost->_value); return _RemoveR(root->_right, key); } delete del; } return true; } //中序遍歷遞歸寫法 void InOrder() { _InOrder(_root); } void _InOrder(Node* root) { if (NULL == root) return; _InOrder(root->_left); cout< _key<<" "; _InOrder(root->_right); } protected: Node* _root; }; void Test() { BSTree t; int a[] = {5, 3, 4, 1, 7, 8, 2, 6, 0, 9}; for (size_t i = 0; i < sizeof(a)/sizeof(a[0]);++i) { t.InsertR(a[i], i); } cout< _key< _key< _key< “C++中如何實(shí)現(xiàn)搜索二叉樹”的內(nèi)容就介紹到這里了,感謝大家的閱讀。如果想了解更多行業(yè)相關(guān)的知識(shí)可以關(guān)注創(chuàng)新互聯(lián)網(wǎng)站,小編將為大家輸出更多高質(zhì)量的實(shí)用文章!
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