105. 从前序与中序遍历序列构造二叉树

 1 /**
 2  * Definition for a binary tree node.
 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 
11 //直接模拟
12 class Solution 
13 {
14 public:
15     TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) 
16     {
17         int n = preorder.size();
18         if(n == 0) return NULL;
19         return build(preorder,0,n - 1,inorder,0,n - 1);
20     }
21 
22     TreeNode* build(vector<int>& preorder,int pb,int pe,vector<int>& inorder, int ib,int ie)
23     {
24         int val = preorder[pb];
25         TreeNode* root = new TreeNode(val);
26 
27         int i = ib;
28         for(;i <= ie;i ++) if(inorder[i] == val) break;
29 
30         //有左子树就是i不能为ib
31         if(i != ib) root->left = build(preorder,pb + 1,pb + i - ib,inorder,ib,i - 1);
32         else root->left = 0;
33         
34         //有右子树就是i不能为ie
35         if(i != ie) root->right= build(preorder,pb + i - ib + 1,pe,inorder,i + 1,ie);
36         else root->right = 0;
37 
38         return root;
39     }
40 
41 };