Leetcode145: Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

/**
* 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* dfs(vector<int>& inorder, int instart, int inend, vector<int>& postorder, int poststart, int postend)
{
if(instart > inend)
return NULL;

TreeNode* root = new TreeNode(0);
root->val = postorder.at(postend);
if(instart == inend)
return root;
int i;
for(i = 0; i <= (inend-instart); i++)
{
if(inorder.at(instart+i) == root->val)
break;
}
root->left = dfs(inorder, instart, instart+i-1, postorder, poststart, poststart+i-1);
root->right = dfs(inorder, instart+i+1, inend, postorder, poststart+i, postend-1);

return root;

}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
if(inorder.size()<=0 || inorder.size()!=postorder.size())
return NULL;
TreeNode* root = new TreeNode(0);
root = dfs(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
return root;
}
};