[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 };