Java深入了解数据结构之二叉搜索树增 插 删 创详解
目录
- ①概念
- ②操作-查找
- ③操作-插入
- ④操作-删除
- 1. cur.left == null
- 2. cur.right == null
- 3. cur.left != null && cur.right != null
- ⑤性能分析
- ⑥完整代码
①概念
二叉搜索树又称二叉排序树,它或者是一棵空树**,或者是具有以下性质的二叉树:
若它的左子树不为空,则左子树上所有节点的值都小于根节点的值
若它的右子树不为空,则右子树上所有节点的值都大于根节点的值
它的左右子树也分别为二叉搜索树
②操作-查找
二叉搜索树的查找类似于二分法查找
public Node search(int key) { Node cur = root; while (cur != null) { if(cur.val == key) { return cur; }else if(cur.val < key) { cur = cur.right; }else { cur = cur.left; } } return null; }
③操作-插入
public boolean insert(int key) { Node node = new Node(key); if(root == null) { root = node; return true; } Node cur = root; Node parent = null; while(cur != null) { if(cur.val == key) { return false; }else if(cur.val < key) { parent = cur; cur = cur.right; }else { parent = cur; cur = cur.left; } } //parent if(parent.val > key) { parent.left = node; }else { parent.right = node; } return true; }
④操作-删除
删除操作较为复杂,但理解了其原理还是比较容易
设待删除结点为 cur, 待删除结点的双亲结点为 parent
1. cur.left == null
1. cur 是 root,则 root = cur.right
2. cur 不是 root,cur 是 parent.left,则 parent.left = cur.right
3. cur 不是 root,cur 是 parent.right,则 parent.right = cur.right
2. cur.right == null
1. cur 是 root,则 root = cur.left
2. cur 不是 root,cur 是 parent.left,则 parent.left = cur.left
3. cur 不是 root,cur 是 parent.right,则 parent.right = cur.left
第二种情况和第一种情况相同,只是方向相反,这里不再画图
3. cur.left != null && cur.right != null
需要使用替换法进行删除,即在它的右子树中寻找中序下的第一个结点(关键码最小),用它的值填补到被删除节点中,再来处理该结点的删除问题
当我们在左右子树都不为空的情况下进行删除,删除该节点会破坏树的结构,因此用替罪羊的方法来解决,实际删除的过程还是上面的两种情况,这里还是用到了搜索二叉树的性质
public void remove(Node parent,Node cur) { if(cur.left == null) { if(cur == root) { root = cur.right; }else if(cur == parent.left) { parent.left = cur.right; }else { parent.right = cur.right; } }else if(cur.right == null) { if(cur == root) { root = cur.left; }else if(cur == parent.left) { parent.left = cur.left; }else { parent.right = cur.left; } }else { Node targetParent = cur; Node target = cur.right; while (target.left != null) { targetParent = target; target = target.left; } cur.val = target.val; if(target == targetParent.left) { targetParent.left = target.right; }else { targetParent.right = target.right; } } } public void removeKey(int key) { if(root == null) { return; } Node cur = root; Node parent = null; while (cur != null) { if(cur.val == key) { remove(parent,cur); return; }else if(cur.val < key){ parent = cur; cur = cur.right; }else { parent = cur; cur = cur.left; } } }
⑤性能分析
插入和删除操作都必须先查找,查找效率代表了二叉搜索树中各个操作的性能。
对有n个结点的二叉搜索树,若每个元素查找的概率相等,则二叉搜索树平均查找长度是结点在二叉搜索树的深度 的函数,即结点越深,则比较次数越多。
但对于同一个关键码集合,如果各关键码插入的次序不同,可能得到不同结构的二叉搜索树:
最优情况下,二叉搜索树为完全二叉树,其平均比较次数为:
最差情况下,二叉搜索树退化为单支树,其平均比较次数为:
⑥完整代码
public class TextDemo { public static class Node { public int val; public Node left; public Node right; public Node (int val) { this.val = val; } } public Node root; /** * 查找 * @param key */ public Node search(int key) { Node cur = root; while (cur != null) { if(cur.val == key) { return cur; }else if(cur.val < key) { cur = cur.right; }else { cur = cur.left; } } return null; } /** * * @param key * @return */ public boolean insert(int key) { Node node = new Node(key); if(root == null) { root = node; return true; } Node cur = root; Node parent = null; while(cur != null) { if(cur.val == key) { return false; }else if(cur.val < key) { parent = cur; cur = cur.right; }else { parent = cur; cur = cur.left; } } //parent if(parent.val > key) { parent.left = node; }else { parent.right = node; } return true; } public void remove(Node parent,Node cur) { if(cur.left == null) { if(cur == root) { root = cur.right; }else if(cur == parent.left) { parent.left = cur.right; }else { parent.right = cur.right; } }else if(cur.right == null) { if(cur == root) { root = cur.left; }else if(cur == parent.left) { parent.left = cur.left; }else { parent.right = cur.left; } }else { Node targetParent = cur; Node target = cur.right; while (target.left != null) { targetParent = target; target = target.left; } cur.val = target.val; if(target == targetParent.left) { targetParent.left = target.right; }else { targetParent.right = target.right; } } } public void removeKey(int key) { if(root == null) { return; } Node cur = root; Node parent = null; while (cur != null) { if(cur.val == key) { remove(parent,cur); return; }else if(cur.val < key){ parent = cur; cur = cur.right; }else { parent = cur; cur = cur.left; } } } }
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