标签：lin tree code into return logs void play bsp
实现splayTree，不过感觉大头都已经写好了，难度不算太大。
part1：zig-zig，类似zig-zag，不过是先要rotate node.parent再rotate node。
private void zigZig(BinaryTreeNode node) { if(node==node.parent.leftChild){ rotateRight(node.parent); rotateRight(node); } else{ rotateLeft(node.parent); rotateLeft(node); } }
part2：splaynode：
private void splayNode(BinaryTreeNode node) { while (node.parent != null) { if(node.parent.parent!=null){ if(node.parent.parent.leftChild==node.parent&&node.parent.leftChild==node||node.parent.parent.rightChild==node.parent&&node.parent.rightChild==node){ zigZig(node); } else { zigZag(node); } } else { zig(node); } } root = node; }
运行结果：
PART I: Testing zigZig() Inserting 1G, 3O, 2O, 5J, 4D, 7B, 6O into Tree 1. Tree 1 is: (((((1G)2O)3O)4D)5J)6O(7B) Skipping the rest of Part I. PART II: Testing splayNode() Calling splayNode() on the unbalanced tree: Inserting 10A, 9B, 8C, 7D, 6E, 5F, 4G, 3H, 2I, 1J tree is: 1J(2I(3H(4G(5F(6E(7D(8C(9B(10A))))))))) Calling find(10) returned A. The tree should be better balanced now: (1J((2I)3H((4G)5F((6E)7D((8C)9B)))))10A Some find() operations on a new tree to test splayNode(): Inserting 3A, 7B, 4C, 2D, 9E, 1F Tree 2 is: 1F((2D(3A((4C)7B)))9E) Calling find(7) returned B. Tree 2 is: (1F((2D)3A(4C)))7B(9E) Calling find(4) returned C. Tree 2 is: ((1F(2D))3A)4C(7B(9E))
标签：lin tree code into return logs void play bsp
原文地址：http://www.cnblogs.com/lyz1995/p/7280809.html
