题目：

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

解题思路：

前序遍历序列的第一个元素肯定为根节点；然后再在中序遍历序列中找到该节点，并以此将中序遍历序列分为左右两部分，然后对左右两部分分别进行递归求解。

代码：

 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     int findIndex(int target, vector<int> &inorder, int begin, int end) {
13         for (int i = begin; i <= end; i++) {
14             if (inorder[i] == target) {
15                 return i;
16             }
17         }
18         return -1;
19     }
20     TreeNode *buildTreeCore(vector<int> &preorder, int pre_begin, int pre_end, vector<int> &inorder, int in_begin, int in_end) {
21         if (pre_begin > pre_end || in_begin > in_end) return NULL;
22
23         TreeNode *root = new TreeNode(preorder[pre_begin]);
24         if (pre_begin == pre_end) {
25             return root;
26         } else {
27             int index = findIndex(preorder[pre_begin], inorder,in_begin, in_end);
28             int left_nodes = index - in_begin;
29             TreeNode *left_root = buildTreeCore(preorder, pre_begin + 1, pre_begin + left_nodes, inorder, in_begin, index - 1);
30             TreeNode *right_root = buildTreeCore(preorder, pre_begin + left_nodes + 1, pre_end, inorder, index + 1, in_end);
31             root->left = left_root;
32             root->right = right_root;
33             return root;
34         }
35     }
36     TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
37         TreeNode *root = buildTreeCore(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
38         return root;
39     }
40 };

转载于:https://www.cnblogs.com/dongguangqing/p/3728246.html

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解题思路：

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