剑指 Offer 07. 重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请构建该二叉树并返回其根节点。

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。

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

示例2

Input: preorder = [-1], inorder = [-1]
Output: [-1]

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int n = 0;
    vector<int>pre;
    vector<int>in;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        n = preorder.size();
        if(n==0)
            return nullptr;
        else
        {
            TreeNode* ans = nullptr;
            pre = preorder;
            in = inorder;
            create_tree(ans,0,n-1,0,n-1);
            return ans;
        }
    }
    void create_tree(TreeNode*& ans,int pre_left, int pre_right,int in_left,int in_right)
    {
        if(pre_left>pre_right)
            return ;
        int i = 0;
        for(i = in_left;i<=in_right;i++)
        {
            if(pre[pre_left] == in[i])
                break;
        }
        ans = new TreeNode(pre[pre_left]);
        create_tree(ans->left,pre_left+1,pre_left+i-in_left,in_left,i-1);
        create_tree(ans->right,pre_left+1+i-in_left,pre_right,i+1,in_right);
    }
};
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此题最重要的就是推导公式，根据前序遍历去找中序遍历的下标，推算出左右子树在前序遍历的范围。