[LeetCode]106. Construct Binary Tree from Inorder and Postorder Traversal

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.

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/** * 和之前使用中序和前序遍历重建的题一样，可以直接使用索引，避免反复创建vector，这里就省略不实现了... *//** * 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.empty() || postorder.empty())            return NULL;        TreeNode* root = new TreeNode(*(postorder.end()-1));        int i = 0;        for(; i < inorder.size(); ++i){            if(inorder[i] == root->val)                break;        }        if(i > 0){            vector<int> in_left = vector<int>(inorder.begin(), inorder.begin()+i);            vector<int> post_left = vector<int>(postorder.begin(), postorder.begin()+i);            root->left = buildTree(in_left, post_left);        }        if(i < inorder.size()-1){            vector<int> in_right = vector<int>(inorder.begin()+i+1, inorder.end());            vector<int> post_right = vector<int>(postorder.begin()+i, postorder.end()-1);            root->right = buildTree(in_right, post_right);        }        return root;    }};