把二元查找树转变成排序的双向链表

1.把二元查找树转变成排序的双向链表

题目：

输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。

要求不能创建任何新的结点，只调整指针的指向。

10

/ \

6 14

/ \ / \

4 8 12 16

转换成双向链表

4=6=8=10=12=14=16。

首先我们定义的二元查找树节点的数据结构如下：

struct BSTreeNode

{

int m_nValue; // value of node

BSTreeNode *m_pLeft; // left child of node

BSTreeNode *m_pRight; // right child of node

};

ANSWER:

This is a traditional problem that can be solved using recursion.

For each node, connect the double linked lists created from left and right child node to form a full list.

/**

* @param root The root node of the tree

* @return The head node of the converted list.

*/

BSTreeNode * treeToLinkedList(BSTreeNode * root) {

BSTreeNode * head, * tail;

helper(head, tail, root);

return head;

}

void helper(BSTreeNode *& head, BSTreeNode *& tail, BSTreeNode *root) {

BSTreeNode *lt, *rh;

if (root == NULL) {

head = NULL, tail = NULL;

return;

}

helper(head, lt, root->m_pLeft);

helper(rh, tail, root->m_pRight);

if (lt!=NULL) {

lt->m_pRight = root;

root->m_pLeft = lt;

} else {

head = root;

}

if (rh!=NULL) {

root->m_pRight=rh;

rh->m_pLeft = root;

} else {

tail = root;

}

}

以上部分为从 v_JULY_v博客转载而来