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.

Analyse: The left part of the last element of the postorder sequence in the inorder sequence is its left subtree, and the right part is the right subtree. The same as Construct Binary Tree From Preorder and Inorder Traversal.

Runtime: 52ms.

/**
 * 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* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size() == 0) return NULL;
        return build(postorder, 0, postorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
    TreeNode* build(vector<int>& postorder, int postLeft, int postRight, vector<int>& inorder, int inLeft, int inRight){
        if(inLeft > inRight) return NULL;
        
        TreeNode* root = new TreeNode(postorder[postRight]);
        if(postLeft == postRight) return root;
        
        int index;
        for(index = 0; index < inorder.size(); index++){
            if(inorder[index] == root->val) break;
        }
        root->left = build(postorder, postLeft, postLeft + index - inLeft - 1, inorder, inLeft, index - 1);
        root->right = build(postorder, postLeft + index - inLeft, postRight - 1, inorder, index + 1, inRight);
        
        return root;
    }
};