剑指offer---重建二叉树

题目：重建二叉树

要求：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {

    }
};

解题代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {
        if(pre.size() != in.size() || pre.size() == 0)
            return nullptr;

        vector<int> leftPre, rightPre;
        vector<int> leftIn, rightIn;

        int rootValue = pre[0];
        TreeNode* root = new TreeNode(rootValue);

        int mid;
        for(int i=0; i<in.size(); i++){
            if(rootValue == in[i]){
                mid = i;
                break;
            }
        }

        for(int i=1; i<pre.size(); i++){
            if(i <= mid)
                leftPre.push_back(pre[i]);
            else
                rightPre.push_back(pre[i]);
        }

        for(int i=0; i<pre.size(); i++){
            if(i < mid)
                leftIn.push_back(in[i]);
            else if(i > mid)
                rightIn.push_back(in[i]);
        }
        root->left = reConstructBinaryTree(leftPre, leftIn);
        root->right = reConstructBinaryTree(rightPre, rightIn);
        return root;
    }
};