剑指offer系列-07 重建二叉树

// 解法一：
class Solution {
public:
    unordered_map<int,int> mp;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.empty()) return nullptr;
        // 利用哈希表加速中序遍历结果中根的查找
        int sz=preorder.size()-1;
        for(int i=0;i<sz+1;i++){
            mp[inorder[i]]=i;
        }
        return dfsBuild(preorder,0,sz,inorder,0,sz);
    }
    TreeNode* dfsBuild(vector<int>& preorder,int ps,int pe,vector<int>& inorder,int is,int ie){
        if(ps>pe) return nullptr;
        int root_val=preorder[ps];
        TreeNode* root=new TreeNode(root_val);
        int root_index_in_inorder=mp[root_val];
        int cnt_left=root_index_in_inorder-is;
        root->left=dfsBuild(preorder,ps+1,ps+cnt_left,inorder,is,root_index_in_inorder-1);
        root->right=dfsBuild(preorder,ps+cnt_left+1,pe,inorder,root_index_in_inorder+1,ie);
        return root;
    }
};