[何海涛] 求二元查找树的镜像
题目：输入一颗二元查找树，将该树转换为它的镜像，即在转换后的二元查找树中，左子树的结点都大于右子树的结点。用递归和循环两种方法完成树的镜像转换。
例如输入：
8
/ \
6 10
/\ /\
5 7 9 11
输出：
8
/ \
10 6
/\ /\
11 9 7 5
递归或者栈模拟
1 #include <iostream> 2 #include <stack> 3 using namespace std; 4 5 struct Node 6 { 7 int val; 8 Node *left; 9 Node *right; 10 Node():left(NULL), right(NULL){} 11 }; 12 13 void convert(Node *root) 14 { 15 if (root == NULL) 16 return; 17 18 convert(root->left); 19 convert(root->right); 20 Node *node = root->left; 21 root->left = root->right; 22 root->right = node; 23 } 24 25 void stackConvert(Node *root) 26 { 27 stack<Node* > s; 28 s.push(root); 29 30 while(!s.empty()) 31 { 32 Node *node = s.top(); 33 s.pop(); 34 35 Node *temp = node->left; 36 node->left = node->right; 37 node->right = temp; 38 39 if (node->left) 40 s.push(node->left); 41 if (node->right) 42 s.push(node->right); 43 } 44 } 45 46 void print(Node *node) 47 { 48 if (node == NULL) 49 return; 50 51 cout << "Node:" << node << " val:" << node->val << " left:" << node->left << " right:" << node->right << endl; 52 53 print(node->left); 54 print(node->right); 55 } 56 57 int main() 58 { 59 Node node[7]; 60 node[0].val = 8; 61 node[0].left = &node[1]; 62 node[0].right = &node[2]; 63 64 node[1].val = 6; 65 node[1].left = &node[3]; 66 node[1].right = &node[4]; 67 68 node[2].val = 10; 69 node[2].left = &node[5]; 70 node[2].right = &node[6]; 71 72 node[3].val = 5; 73 node[4].val = 7; 74 node[5].val = 9; 75 node[6].val = 11; 76 77 //convert(&node[0]); 78 stackConvert(&node[0]); 79 80 print(&node[0]); 81 cout << endl; 82 83 Node a[2]; 84 a[0].val = 1; 85 a[0].left = &a[1]; 86 87 a[1].val = 2; 88 89 //convert(&a[0]); 90 stackConvert(&a[0]); 91 print(&a[0]); 92 }
