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.

思路：递归。主要是注意调用时起始和终止的index。

 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         TreeNode* root=process(preorder, 0,preorder.size(),inorder,0,inorder.size());
14         return root;
15     }
16    TreeNode*  process(vector<int> &preorder, int s1, int e1, vector<int> &inorder, int start, int end)
17     {
18         if(start>=end)
19         {
20             return NULL;
21         }
22         int root_val=preorder[s1];
23         int inorder_index=start;
24         while(inorder_index<end)
25         {
26             if(inorder[inorder_index]==root_val)
27             break;
28             else inorder_index++;
29         }
30         int count=inorder_index-start;//做子树节点数量
31         TreeNode* left=process(preorder, s1+1,s1+count+1,inorder,start,inorder_index);
32         TreeNode* right=process(preorder,s1+count+1, e1,inorder, inorder_index+1,end);
33         TreeNode* root=new TreeNode(root_val);
34         root->left=left;
35         root->right=right;
36         return root;
37     }
38 };

 

posted @ 2014-07-09 13:27  Hicandyman  阅读(100)  评论(0)  编辑  收藏  举报