LeetCode Construct Binary Tree from Preorder and Inorder Traversal

题目：

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

Note:

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

分析参考：http://leetcode.com/2011/04/construct-binary-tree-from-inorder-and-preorder-postorder-traversal.html

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
TreeNode *root = createTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
return root;
}
private:
TreeNode *createTree(vector<int> &preorder, int preBeg, int preEnd, vector<int> &inorder, int inBeg, int inEnd) {
if(preBeg > preEnd)
return NULL;
int rootVal = preorder[preBeg];
TreeNode *root = new TreeNode(rootVal);
//找到rootVal在中序中的索引
int index;
for(int i = inBeg; i <= inEnd; i++) {
if(inorder[i] == rootVal) {
index = i;
break;
}
}
int length = index - inBeg;
root->left = createTree(preorder, preBeg+1, preBeg+length, inorder, inBeg, index-1);
root->right = createTree(preorder, preBeg+length+1, preEnd, inorder, index+1, inEnd);
return root;
}
};