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.

For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7

解析：
根据中序和后序遍历顺序生成二叉树。题目类似于LeetCode 105. Construct Binary Tree from Preorder and Inorder Traversal
代码如下：

/**
 * 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) {
        return buildTree(postorder, 0, postorder.size()-1, inorder, 0, inorder.size()-1);
    }
    TreeNode* buildTree(vector<int>& postorder, int pLeft, int pRight, vector<int>& inorder, int iLeft, int iRight)
    {
        if(pLeft > pRight || iLeft > iRight) return NULL;
        int i =0;
        for(i = iLeft; i <= iRight; ++i)
        {
            if(postorder[pRight] == inorder[i])
                break;
        }
        TreeNode *cur = new TreeNode(postorder[pRight]);
        cur->left = buildTree(postorder, pLeft, pLeft+i-iLeft-1, inorder, iLeft, i-1);
        cur->right = buildTree(postorder, pLeft+i-iLeft, pRight-1, inorder, i+1, iRight);
        return cur;
    }
};