Leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

原题：
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, givenpreorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]Return the following binary tree: 3
/ \
9 20
/ \
15 7解决方法：
- 前序遍历的第一个节点是根节点。
- 根据根节点把中序遍历的点分成左右两部分。
- 根据中序遍历左右部分的长度将前序遍历也分割成左右两部分。
这样递归循环即可求出最后的树。

代码：

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder, int pre_start, int pre_stop, int in_start, int in_stop){
if (pre_stop < pre_start){
return NULL;
}

TreeNode* root = new TreeNode(preorder[pre_start]);
int pos;
for(int i = in_start; i <= in_stop;i++){
if (preorder[pre_start] == inorder[i]){
pos = i;
break;
}
}
root->left = buildTree(preorder, inorder, pre_start + 1, pre_start + pos - in_start, in_start, pos - 1);
root->right = buildTree(preorder, inorder, pre_stop - in_stop + pos + 1, pre_stop, pos + 1, in_stop);
return root;
}

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {

return buildTree(preorder, inorder, 0, preorder.size()-1, 0, inorder.size() - 1);
}