# My C++ Solution

• ``````class Solution {

public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {

return helper(inorder,0,inorder.size(),postorder,0,postorder.size());
}
private:
TreeNode* helper(vector<int>& inorder,int i,int j,vector<int>& postorder,int ii,int jj)
{
// 每次取postorder的最后一个值mid，将其作为树的根节点
// 然后从inroder中找到mid，将其分割成为两部分，左边作为mid的左子树，右边作为mid的右子树
// tree:     8 4 10 3 6 9 11
// Inorder   [3 4 6] 8 [9 10 11]
// postorder [3 6 4]   [9 11 10] 8

if(i >= j || ii >= jj)
return NULL;

int mid = postorder[jj - 1];

auto f = find(inorder.begin() + i,inorder.begin() + j,mid);

int dis = f - inorder.begin() - i;

TreeNode* root = new TreeNode(mid);
root -> left = helper(inorder,i,i + dis,postorder,ii,ii + dis);
root -> right = helper(inorder,i + dis + 1,j,postorder,ii + dis,jj - 1);

return root;

}
};``````

• Super clear solution!
I suggest you can replace
`find(inorder.begin() + i,inorder.begin() + j,mid);`
to
`lower_bound(inorder.begin() + i,inorder.begin() + j,mid);`
since inOrder array is in order, can use binary search.

• Oh, sorry, it is not ture that inorder array is in order, forget about my suggestion.

• My solution is similar but only uses iterator f to do the math. The use of dis inspired me and I made my solution a lot more clear. Thanks!

• Why binary search is not working here?

• It's not binary search tree.

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