[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.

 

方法:在inorder中寻找postorder的最后一个,然后左右递归,注意处理好各个low,high边界

 1 class Solution
 2 {
 3     public:
 4         TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
 5         {
 6             return buildTree(postorder, 0, postorder.size()-1,
 7                               inorder, 0, inorder.size()-1);
 8         }
 9 
10         TreeNode *buildTree(vector<int> &postorder, int low1, int high1,
11                             vector<int> &inorder, int low2, int high2)
12         {
13             //cout << "==============" <<endl;
14             //cout << "low1 = \t" << low1 <<endl;
15             //cout << "high1= \t" << high1 <<endl;
16             //cout << "low2 = \t" << low2 <<endl;
17             //cout << "high2= \t" << high2 <<endl;
18             if(low1 > high1 || low2> high2)
19                 return NULL;
20 
21             TreeNode * p = new TreeNode(postorder[high1]);
22             if(low1 == high1)
23             {   
24                 return p;
25             }   
26             int index = 0;
27             for(index = low2; index < high2; index++)
28             {   
29                 if(inorder[index] == postorder[high1])
30                     break;
31             }   
32             //cout << "index= \t" << index<<endl;
33 
34             if(index != low2)
35                 p->left = buildTree(postorder, low1,(low1) + (index-1-low2), inorder, low2, index-1);
36             if(index != high2)
37                 p->right = buildTree(postorder, (high1-1) - (high2-index-1) ,high1-1, inorder, index+1, high2);
38 
39             return p;
40         }
41 } ;

 

posted @ 2014-06-27 22:40  穆穆兔兔  阅读(160)  评论(0编辑  收藏  举报