LeetCode105

重建二叉树

分析

这道题难道是不难，主要是我一边写一边想事情，然后写完提交就过了。。。。直接在页面上写的，编译也是一次就过了
记录一下

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return getAns(preorder,inorder);
    }
    TreeNode * getAns(vector<int>& preorder, vector<int>& inorder){
        if(!preorder.size())return NULL;
        if(preorder.size()==1){
            return new TreeNode(preorder[0]);
        }
        TreeNode * rt=new TreeNode(preorder[0]);
        vector<int>lpre,rpre;
        vector<int>lmid,rmid;
        int preFlag=0,midFlag=0;
        for(int i=0;i<inorder.size();i++){
            if(inorder[i]==preorder[0]){
                midFlag=1;
                continue;
            }
            if(!midFlag){
                lmid.push_back(inorder[i]);
            }else{
                rmid.push_back(inorder[i]);
            }
        }
        int lsiz=lmid.size();
        for(int i=1;i<preorder.size();i++){
            if(i<1+lsiz){
                lpre.push_back(preorder[i]);
            }else{
                rpre.push_back(preorder[i]);
            }
        }
        rt->left=getAns(lpre,lmid);
        rt->right=getAns(rpre,rmid);
        return rt;
    }
};