105. 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

同第106题，只是树根是先序序列的第一个元素

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
typedef vector<int>::iterator Iter;
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
return buildTreeRecur(inorder.begin(), inorder.end(), preorder.begin(), preorder.end());
}
TreeNode *buildTreeRecur(Iter istart, Iter iend, Iter pstart, Iter pend)
{
if(istart == iend)return NULL;
int rootval = *pstart;
Iter iterroot = find(istart, iend, rootval);
TreeNode *res = new TreeNode(rootval);
res->left = buildTreeRecur(istart, iterroot, pstart+1, pstart+1+(iterroot-istart));
res->right = buildTreeRecur(iterroot+1, iend, pstart+1+(iterroot-istart), pend);
return res;
}
};