Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
class Solution { public: TreeNode *buildTree(vector<int>& inorder, int in_left,int in_right, vector<int>& postorder, int post_left, int post_right){ if(in_right < in_left || post_right < post_left) return NULL; TreeNode *root = new TreeNode(postorder[post_right]); int index = in_left; for( ; index <= in_right; ++ index ) if(inorder[index] == postorder[post_right]) break; int right_cnt = in_right-index; root->left = buildTree(inorder,in_left,index-1,postorder,post_left,post_right-right_cnt-1); root->right = buildTree(inorder,index+1,in_right,postorder, post_right-right_cnt,post_right-1); return root; } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { if(inorder.size() == 0) return NULL; else return buildTree(inorder,0,inorder.size()-1, postorder,0,postorder.size()-1); } };