给定一棵树的中序遍历与后序遍历，依据此构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
中序遍历 = [9,3,15,20,7]
后序遍历 = [9,15,7,20,3]
返回如下的二叉树：
3
/ \
9 20
/ \
15 7
详见：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
int is=inorder.size();
if(is==0||inorder.empty())
{
return nullptr;
}
int val=postorder[is-1];
TreeNode* root=new TreeNode(val);
vector<int> in_left,in_right,post_left,post_right;
int p=0;
for(;p<is;++p)
{
if(inorder[p]==val)
{
break;
}
}
for(int i=0;i<is;++i)
{
if(i<p)
{
in_left.push_back(inorder[i]);
post_left.push_back(postorder[i]);
}
else if(i>p)
{
in_right.push_back(inorder[i]);
post_right.push_back(postorder[i-1]);
}
}
root->left=buildTree(in_left,post_left);
root->right=buildTree(in_right,post_right);
return root;
}
};