给出一棵BST，将其转换为双向链表，left保存前节点，right保存后节点。
例如：
4
/ \
1 7
/ \ / \
0 2 5 8
\ \ \
3 6 9
变为：0<->1<->2<->3<->4<->5<->6<->7<->8<->9.
解答：这里我们可以看到一个节点的左子树的最后节点就是本节点的前驱，右子树的最左节点就是后继。于是需要一个函数能够返回这样的需求，也就有了一个可以返回本棵树的左右节点的函数。
同时，在细节上进行一些处理。
1 #include <iostream> 2 using namespace std; 3 4 struct Node 5 { 6 int val; 7 Node *left; 8 Node *right; 9 Node():left(NULL), right(NULL){} 10 }; 11 12 struct LeftRightNode 13 { 14 Node *left; 15 Node *right; 16 LeftRightNode(){} 17 LeftRightNode(Node *l, Node *r):left(l), right(r){} 18 }; 19 20 LeftRightNode convert(Node *node) 21 { 22 if (node == NULL) 23 { 24 LeftRightNode ret; 25 return LeftRightNode(NULL, NULL); 26 } 27 28 LeftRightNode leftTree = convert(node->left); 29 LeftRightNode rightTree = convert(node->right); 30 31 node->left = leftTree.right; 32 if (leftTree.right) 33 leftTree.right->right = node; 34 35 node->right = rightTree.left; 36 if (rightTree.left) 37 rightTree.left->left = node; 38 39 Node *leftNode = leftTree.left ? leftTree.left : node; 40 Node *rightNode = rightTree.right ? rightTree.right : node; 41 return LeftRightNode(leftNode, rightNode); 42 } 43 44 Node *convert2List(Node *head) 45 { 46 LeftRightNode ret = convert(head); 47 48 return ret.left; 49 } 50 51 void print(Node *node) 52 { 53 while(node) 54 { 55 cout << "Node val:" << node->val << " address:" << node << " left:" << node->left << " right:" << node->right << endl; 56 node = node->right; 57 } 58 } 59 60 int main() 61 { 62 Node node[10]; 63 for(int i = 0; i < 10; i++) 64 node[i].val = i; 65 66 node[4].left = &node[1]; 67 node[4].right = &node[7]; 68 69 node[1].left = &node[0]; 70 node[1].right = &node[2]; 71 72 node[2].right = &node[3]; 73 74 node[7].left = &node[5]; 75 node[7].right = &node[8]; 76 77 node[5].right = &node[6]; 78 79 node[8].right = &node[9]; 80 81 Node *head = convert2List(&node[4]); 82 83 print(head); 84 }