1、概念
二叉搜索树性质:设x是二叉搜索树的一个节点,那么:
a) 对x左子树中任意节点y, y.key < x.key
b) 对x右子树种任意节点y, y.key >= x.key
2、数据结构
1 struct TreeNode 2 { 3 TreeNode(int key): left(NULL), right(NULL), parent(NULL), key(key) {}; 4 TreeNode* left; 5 TreeNode* right; 6 TreeNode* parent; 7 int key; 8 }; 9 10 class Tree 11 { 12 public: 13 Tree(): m_root(NULL) {}; 14 TreeNode* minimum(TreeNode* ); 15 TreeNode* maximum(TreeNode* ); 16 void creat(int keys[], int n); 17 int insert(int key); 18 int remove(int key); 19 TreeNode* search(int key); 20 void walk(TreeNode *root); 21 TreeNode* get_root() const; 22 23 protected: 24 inline TreeNode* node_creat(int key); 25 void transplant(TreeNode* x, TreeNode *y); 26 27 private: 28 TreeNode* m_root; 29 };
3、算法
3.1 遍历
由二叉搜索树的性质,中序遍历的结果为keys的有序排列
1 void Tree::walk(TreeNode *root) 2 { 3 if (root == NULL) 4 return; 5 walk(root->left); 6 cout << root->key << " "; 7 walk(root->right); 8 }
3.2 最大/最小值
1 TreeNode* Tree::minimum(TreeNode *root) 2 { 3 TreeNode *visit = root; 4 if (visit == NULL) 5 return NULL; 6 while (visit->left != NULL) 7 visit = visit->left; 8 return visit; 9 } 10 11 TreeNode* Tree::maximum(TreeNode* root) 12 { 13 TreeNode *visit = root; 14 if (visit == NULL) 15 return NULL; 16 while (visit->right != NULL) 17 visit = visit->right; 18 return visit; 19 }
3.3 查找
1 TreeNode* Tree::search(int key) 2 { 3 TreeNode* visit = m_root; 4 5 while (visit != NULL) 6 { 7 if (key == visit->key) 8 return visit; 9 if (key < visit->key) 10 visit = visit->left; 11 else 12 visit = visit->right; 13 } 14 15 return visit; 16 }
3.4 插入
插入时,必然是插入到某个叶子节点上,而不是两节点之间。
先查找到插入位置,然后插入。
1 int Tree::insert(int key) 2 { 3 TreeNode* z = new TreeNode(key); 4 TreeNode* x = m_root; 5 TreeNode* y = NULL; 6 while (x != NULL) 7 { 8 y = x; 9 if (z->key < x->key) 10 x = x->left; 11 else 12 x = x->right; 13 } 14 z->parent = y; 15 if (y == NULL) // tree is empty 16 m_root = z; 17 else if (z->key < y->key) 18 y->left = z; 19 else 20 y->right = z; 21 22 return 0; 23 }
3.5 删除
删除时,情况较复杂,对于待删除节点z:
IF z.left is NULL
replace z by z.right
ELSE IF z.right is NULL
replace z by z.left
ELSE // z 包含左右子树
找出z的后继y,即key大于z.key的最小节点, y在z的右子树中且y没有左子树
IF y.parent is z
replace z by y, 留下y.right
ELSE
replace z.right by y,
replace z by y
1 // replace x by node y 2 void Tree::transplant(TreeNode* x, TreeNode *y) 3 { 4 if (x->parent == NULL) // x is root 5 m_root = y; 6 else if (x->parent->left == x) 7 x->parent->left = y; 8 else 9 x->parent->right = y; 10 11 if (y != NULL) 12 y->parent = x->parent; 13 } 14 15 int Tree::remove(int key) 16 { 17 TreeNode* y = NULL; 18 TreeNode* z = search(key); 19 if (z == NULL) 20 return -1; 21 22 if (z->left == NULL) 23 transplant(z, z->right); 24 else if (z->right == NULL) 25 transplant(z, z->left); 26 else 27 { 28 y = minimum(z->right); 29 if (y->parent != z) 30 { 31 transplant(y, y->right); // replace y by right child 32 // let y be parent of z->right 33 y->right = z->right; 34 y->right->parent = y; 35 } 36 transplant(z, y); 37 y->left = z->left; 38 y->left->parent = y; 39 } 40 41 return 0; 42 }
tips:源码
时间: 2024-10-30 16:39:55