[Leetcode] 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.

 

没什么好说的，了解原理即可，使用递归。

 

 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, int a, int b, int n) {
13         TreeNode *root = NULL;
14         if (n < 1) {
15             return root;
16         }
17         root = new TreeNode(preorder[a]);
18         bool flag = false;
19         int i;
20         for (i = 0; i < n; ++i) {
21             if (inorder[b+i] == preorder[a]) {
22                 break;
23             }
24         }
25         root->left = buildTree(preorder, inorder, a+1, b, i);
26         root->right = buildTree(preorder, inorder, a+i+1, b+i+1, n-1-i);
27         return root;
28     }
29 
30     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
31         int n = preorder.size();
32         return buildTree(preorder, inorder, 0, 0, n);
33         
34     }
35 };

 

posted @ 2014-03-28 14:43  Eason Liu  阅读(231)  评论(0)  编辑  收藏  举报