[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

树的递归

 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     TreeNode *createTree(vector<int> &preorder, int preBeg, int preEnd, vector<int> &inorder, int inBeg, int inEnd)
13     {
14         if (preBeg > preEnd)
15             return NULL;
16             
17         int root = preorder[preBeg];
18         int index;
19         for(int i = inBeg; i <= inEnd; i++)
20             if (root == inorder[i])
21             {
22                 index = i;
23                 break;
24             }
25         
26         int len = index - inBeg;
27         TreeNode *left = createTree(preorder, preBeg + 1, preBeg + len, inorder, inBeg, index - 1);
28         TreeNode *right = createTree(preorder, preBeg + len + 1, preEnd, inorder, index + 1, inEnd);
29         
30         TreeNode *node = new TreeNode(root);
31         
32         node->left = left;
33         node->right = right;
34         
35         return node;        
36     }
37     
38     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
39         // Start typing your C/C++ solution below
40         // DO NOT write int main() function
41         if (preorder.size() == 0)
42             return NULL;
43             
44         TreeNode *head = createTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
45         
46         return head;
47     }
48 };