[LeetCode] 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.

递归构造

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode *createTree(vector<int> &inorder, int inBeg, int inEnd, vector<int> &postorder, int postBeg, int postEnd)
13     {
14         if (inBeg > inEnd)
15             return NULL;
16             
17         int root = postorder[postEnd];
18         
19         int index;
20         
21         for(int i = inBeg; i <= inEnd; i++)
22             if (inorder[i] == root)
23             {
24                 index = i;
25                 break;
26             }
27             
28         int len = index - inBeg;
29         TreeNode *left = createTree(inorder, inBeg, index - 1, postorder, postBeg, postBeg + len - 1);
30         TreeNode *right = createTree(inorder, index + 1, inEnd, postorder, postBeg + len, postEnd - 1);
31         
32         TreeNode *node = new TreeNode(root);
33         node->left = left;
34         node->right = right;
35         
36         return node;
37     }
38     
39     TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
40         // Start typing your C/C++ solution below
41         // DO NOT write int main() function
42         if (inorder.size() == 0)
43             return NULL;
44             
45         TreeNode *head = createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);
46         
47         return head;
48     }
49 };