Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

Solution: Recursion.

 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 *buildTree(vector<int> &preorder, vector<int> &inorder) {
13         return buildTreeRe(preorder.begin(), inorder.begin(), preorder.size());
14     }
15     
16     TreeNode* buildTreeRe(vector<int>::iterator preorder, vector<int>::iterator inorder, int N) 
17     {
18         if(N <= 0) return NULL;
19         vector<int>::iterator it = find(inorder, inorder + N, *preorder);
20         int pos = it - inorder;
21         TreeNode* root = new TreeNode(*preorder);
22         root->left = buildTreeRe(preorder+1, inorder, pos);
23         root->right = buildTreeRe(preorder+pos+1, inorder+pos+1, N-1-pos);
24         return root;
25     }
26 };

 

posted @ 2014-04-20 02:35  beehard  阅读(153)  评论(0)  编辑  收藏  举报