剑指Offer——重建二叉树

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int Search(vector<int> &vin, int vinLeft, int vinRight, int key) {
13         int i = (vinLeft + vinRight) >> 1;
14         int j = i + 1;
15         while(i >= vinLeft && j <= vinRight) {  // 从中间往两边查找比较快能找到要找的值的位置
16             if(vin[i] == key) return i;
17             if(vin[j] == key) return j;
18             i--;
19             j++;
20         }
21         return vinLeft;
22     }
23     TreeNode* myReConstructBinaryTree(vector<int> &pre, int preLeft, int preRight, vector<int> &vin, int vinLeft, int vinRight) {
24         if(preLeft > preRight || vinLeft > vinRight) return NULL;
25         TreeNode* t = new TreeNode(pre[preLeft]);
26         int k = Search(vin, vinLeft, vinRight, pre[preLeft]);   // vin中k位置前的肯定是左子树的各个节点的值，k位置后肯定是右子树的各个节点的值
27         t->left = myReConstructBinaryTree(pre, preLeft + 1, preLeft + k - vinLeft, vin, vinLeft, k - 1);
28         t->right = myReConstructBinaryTree(pre, preLeft + k - vinLeft + 1, preRight, vin, k + 1, vinRight);
29         return t;
30     }
31     TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {
32         int Size = pre.size();
33         return myReConstructBinaryTree(pre, 0, Size - 1, vin, 0, Size - 1);
34     }
35 };