leetcode

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 binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    TreeNode *buildTree(std::vector<int> &inorder, std::vector<int> &postorder) {
		return buildBinTree(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);
	}
private:
	TreeNode *buildBinTree(std::vector<int> &inorder,int inStart,int inEnd,std::vector<int> &postorder,int postStart,int postEnd)
	{
		if(inStart > inEnd || postStart > postEnd) return NULL;
		TreeNode *root = new TreeNode(postorder[postEnd]);
		int rootIndex = 0;
		for (int i = inStart; i <= inEnd; i++)
		{
			if(inorder[i] == root->val)
			{
				rootIndex = i;
				break;
			}
		}
		int len = rootIndex - inStart;
		root->left = buildBinTree(inorder,inStart,rootIndex-1,postorder,postStart,postStart+len-1);
		root->right= buildBinTree(inorder,rootIndex+1,inEnd,postorder,postStart+len,postEnd-1);
		return root;
	}
};