根据一棵树的前序遍历与中序遍历构造二叉树。

注意:

你可以假设树中没有重复的元素。

例如,给出

前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树:

3 / \ 9 20 / \ 15 7

class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder)
{
if(preorder.size() == 0)
return NULL;
if(preorder.size() == 1)
return new TreeNode(preorder[0]);
int iroot = preorder[0];
int ipos = 0;
for(int i = 0; i < inorder.size(); i++)
{
if(inorder[i] == iroot)
{
ipos = i;
break;
}
}
TreeNode *root = new TreeNode(iroot);
vector<int> v1(preorder.begin() + 1, preorder.begin() + ipos + 1);
vector<int> v2(inorder.begin(), inorder.begin() + ipos);
vector<int> v3(preorder.begin() + ipos + 1, preorder.end());
vector<int> v4(inorder.begin() + ipos + 1, inorder.end());
root ->left = buildTree(v1, v2);
root ->right = buildTree(v3, v4);
return root;
}
};

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