TreeNode* buildTreeHelper(unordered_map<int, int>& inorder_map, int is, int ie, vector<int>& postorder, int ps, int pe){
if(is > ie || ps > pe){return NULL;}
int root_val = postorder[pe];
TreeNode *root = new TreeNode(root_val);
int inIndex = inorder_map.find(root_val) -> second;
int dis = inIndex - is;
root -> left = buildTreeHelper(inorder_map, is, is + dis - 1, postorder, ps, ps + dis - 1);
root -> right = buildTreeHelper(inorder_map, is + dis + 1, ie, postorder, ps + dis, pe - 1/*last is root*/);
return root;
}
TreeNode* buildTreeHelper(vector<int>& post, int is, int ie, int ps, int pe, unordered_map<int, int>& inorder_map) {
if (is > ie || ps > pe) {
return NULL;
}
int root_val = post[pe];
TreeNode* root = new TreeNode(root_val);
int i = inorder_map.find(root_val)->second;
// number of nodes in left subtree
int l = i-is;
root->left = buildTreeHelper(post, is, is+l-1, ps, ps+l-1, inorder_map);
root->right = buildTreeHelper(post, is+l+1, ie, ps+l, pe-1, inorder_map);
return root;
}
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
// Note: The Solution object is instantiated only once and is reused by each test case.
if (inorder.size() == 0) return NULL;
return buildTreeHelper(inorder, 0 ,inorder.size()-1, postorder, 0, postorder.size()-1);
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder){
unordered_map<int, int> inorder_map;
// we need a map to look up the position of root in inorder, so
// that we can divide the tree into separate subtrees,
// reduces the complexity from n^2 to n assuming good hashing by unodered_map
for(int i = 0; i < inorder.size(); i ++){
inorder_map[inorder[i]] = i;
}
return buildTreeHelper(inorder_map, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
}
TreeNode *buildTreeHelper(vector<int> &inorder, int istart, int iend, vector<int> &postorder, int pstart, int pend){
if (istart == iend || pstart == pend) return new TreeNode(inorder[istart]);
if (istart > iend || pstart > pend) return NULL;
//root is postorder[pend]
TreeNode *root = new TreeNode(postorder[pend]);
// count left child tree node.
int numLeftChild = 0;
for (int i = istart; i<=iend; i++){
if (inorder[i] != postorder[pend]) numLeftChild++;
else break;
}
// now build left,right child recursively
root->left = buildTreeHelper(inorder, istart, istart + numLeftChild - 1, postorder, pstart,pstart+numLeftChild-1);
root->right = buildTreeHelper(inorder, istart + numLeftChild+1, iend, postorder, pstart + numLeftChild,pend-1);
return root;
}
