减治法在查找算法中的应用

二叉查找树的查找与插入：

二叉排序树或者是一棵空树，或者是具有下列性质的二叉树：

（1）若左子树不空，则左子树上所有结点的值均小于或等于它的根节点的值；

（2）若右子树不空，则右子树上所有结点的值均大于或等于它的根结点的值；

（3）左、右子树也分别为二叉排序树。

对于二叉查找树这里我们介绍查找、插入和删除操作：

这些操作会将问题的规模变成一个更小的二叉树，也运用到了减治法的思想。

查找思路：如果树为空，直接返回，查找失败。反之，将查找值key与根root的值比较，如果相等，则root即为所找；若key < root.v，则继续在左子树查找；若key > root.v，则继续在右子树查找。

public static Node find(int key) {if (root == null) {System.out.println("The tree is empty!");return null;}Node cur = root;while (cur.v != key) {if (key < cur.v) {cur = cur.l;}else {cur = cur.r;}if (cur == null) {return null;}}return cur;

插入思路：基本思想和查找操作类似，如果为空树，直接返回，插入节点即为根节点；否则，比较根节点的值与插入节点的值。注意用parent记录遍历到最后的cur值。

public static Node find(int key) {if (root == null) {System.out.println("The tree is empty!");return null;}Node cur = root;while (cur.v != key) {if (key < cur.v) {cur = cur.l;}else {cur = cur.r;}if (cur == null) {return null;}}return cur;
}

删除思路：删除操作是二叉树操作中最复杂的问题。对于二叉树的删除有一下三种情况：

（1）删除叶子结点。可直接删除，不会影响其他

（2）删除结点有且只有一侧孩子结点。孩子结点覆盖待删除结点，删除孩子结点

（3）删除结点下既有左孩子，又有右孩子。在待删除结点的右孩子下，找到v值最小的，用这个结点覆盖要删除的结点。（这里是因为中序遍历结点的后继结点一定是在右子树中v值最小的结点）

public static boolean delete(int key) {Node cur = root;Node parent = root;boolean hasLeft = true;while (cur != null && cur.v != key) {parent = cur;if (key < cur.v) {cur = cur.l;hasLeft = true;} else {cur = cur.r;hasLeft = false;}}if (cur == null) {return false;}if (cur.l == null && cur.r == null) {/*** 要删除的节点为叶子节点，直接删除* */if (cur == root) {root = null;}if (hasLeft) {parent.l = null;} else {parent.r = null;}} else if (cur.r == null) {/*** 要删除的节点只有左孩子* */if (cur == root) {root = cur.l;}if (hasLeft) {parent.l = cur.l;} else {parent.r = cur.l;}} else if (cur.l == null) {/*** 要删除的节点只有右孩子* */if (cur == root) {root = cur.r;}if (hasLeft) {parent.l = cur.r;}else {parent.r = cur.r;}} else {/*** 要删除的节点既有左孩子又有右孩子* 思路：用待删除节点右子树中的v值最小的结点来替代要删除的节点,然后删除右子树中结点* 右子树中v值最小的节点一定没有左子树,所以删除的这个结点一定是属于叶子节点或只有右子树的节点* */Node directPostNode = getPost(cur);cur.v = directPostNode.v;}return true;
}
private static Node getPost(Node delNode) {Node parent = delNode;Node dir = delNode;Node cur = delNode.r;while (cur != null) {parent = dir;dir = cur;cur = cur.l;}if (dir != delNode.r) {//从树中删除此直接后继节点parent.l = dir.r;dir.r = null;}return dir;
}

完整代码如下：

class Node {int v;Node l;Node r;public Node(int v) {this.v = v;}
}
public class Main {public static Node root;public static Node find(int key) {if (root == null) {System.out.println("The tree is empty!");return null;}Node cur = root;while (cur.v != key) {if (key < cur.v) {cur = cur.l;}else {cur = cur.r;}if (cur == null) {return null;}}return cur;}public static void insert(Node node) {if (root == null) {root = node;return;}Node cur = root;Node parent = root;boolean hasLeft = true;while (cur != null) {parent = cur;if (node.v > cur.v) {cur = cur.r;hasLeft = true;} else {cur = cur.l;hasLeft = false;}}if (hasLeft) {parent.l = node;} else {parent.r = node;}}public static boolean delete(int key) {Node cur = root;Node parent = root;boolean hasLeft = true;while (cur != null && cur.v != key) {parent = cur;if (key < cur.v) {cur = cur.l;hasLeft = true;} else {cur = cur.r;hasLeft = false;}}if (cur == null) {return false;}if (cur.l == null && cur.r == null) {/*** 要删除的节点为叶子节点，直接删除* */if (cur == root) {root = null;}if (hasLeft) {parent.l = null;} else {parent.r = null;}} else if (cur.r == null) {/*** 要删除的节点只有左孩子* */if (cur == root) {root = cur.l;}if (hasLeft) {parent.l = cur.l;} else {parent.r = cur.l;}} else if (cur.l == null) {/*** 要删除的节点只有右孩子* */if (cur == root) {root = cur.r;}if (hasLeft) {parent.l = cur.r;}else {parent.r = cur.r;}} else {/*** 要删除的节点既有左孩子又有右孩子* 思路：用待删除节点右子树中的v值最小的结点来替代要删除的节点,然后删除右子树中结点* 右子树中v值最小的节点一定没有左子树,所以删除的这个结点一定是属于叶子节点或只有右子树的节点* */Node directPostNode = getPost(cur);cur.v = directPostNode.v;}return true;}private static Node getPost(Node delNode) {Node parent = delNode;Node dir = delNode;Node cur = delNode.r;while (cur != null) {parent = dir;dir = cur;cur = cur.l;}if (dir != delNode.r) {//从树中删除此直接后继节点parent.l = dir.r;dir.r = null;}return dir;}public static void preorder(Node node) {System.out.print(node.v + " ");if (node.l != null)preorder(node.l);if (node.r != null)preorder(node.r);}public static void main(String[] args) {/*** 插入* */insert(new Node(20));insert(new Node(10));insert(new Node(30));/*** 查找* */System.out.println(find(20));/*** 删除* */delete(20);/*** 前序遍历* */preorder(root);}
}