Leetcode #106 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.

Difficulty:Medium

/**
* 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* build(vector<int>& postorder, vector<int>& inorder,int pStart,int pEnd,int iStart, int iEnd){
//cout<<pEnd<<endl;
TreeNode* root;
if(pEnd<pStart)
return NULL;
int mid = postorder[pEnd];
root = new TreeNode(mid);
if(pEnd == pStart)
return root;
int p_l_s,p_l_e,p_r_s,p_r_e;
int i_l_s,i_l_e,i_r_s,i_r_e;
i_l_s = iStart;
i_l_e = iStart;
while(inorder[i_l_e]!=mid)
i_l_e++;
i_l_e--;
p_l_s = pStart;
p_l_e = p_l_s + (i_l_e - i_l_s);
p_r_s = p_l_e+1;
p_r_e = pEnd-1;
i_r_s = i_l_e + 2;
i_r_e = iEnd;
root->left = build(postorder,inorder,p_l_s,p_l_e,i_l_s,i_l_e);
root->right = build(postorder,inorder,p_r_s,p_r_e,i_r_s,i_r_e);
return root;
}
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
if(postorder.size()==0)
return NULL;
return build(postorder,inorder,0,postorder.size()-1,0,postorder.size()-1);

}
};