重建二叉树
https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=196&tqId=37109&rp=1&ru=/exam/oj&qru=/exam/oj&sourceUrl=%2Fexam%2Foj%3Ftab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196%26page%3D1%26search%3D%25E9%2587%258D%25E5%25BB%25BA%25E4%25BA%258C%25E5%258F%2589%25E6%25A0%2591&difficulty=undefined&judgeStatus=undefined&tags=&title=%E9%87%8D%E5%BB%BA%E4%BA%8C%E5%8F%89%E6%A0%91 先上代码
class Solution {
public:
TreeNode* _reConstructBinaryTree(vector &pre, int pre_start, int pre_end, vector &vin, int vin_start, int vin_end)
{
if(pre_start > pre_end || vin_start > vin_end)
{
return nullptr;
}
TreeNode* root = new TreeNode(pre[pre_start]);
for(auto i = vin_start ; i<= vin_end;++i)
{
if(pre[pre_start] == vin[i])
{
root->left = _reConstructBinaryTree(pre, pre_start + 1 ,pre_start + 1 + i - vin_start -1 , vin,vin_start ,i -1);
root->right = _reConstructBinaryTree(pre, pre_start+ i - vin_start +1,pre_end , vin, i + 1,vin_end);
break;
}
}
return root;
}
TreeNode* reConstructBinaryTree(vector pre,vector vin) {
if(pre.empty()|| vin.empty() || pre.size() != vin.size()) return nullptr;
return _reConstructBinaryTree(pre,0,pre.size() -1,vin,0,vin.size() - 1);
}
}; 列子1：
输入：
[1,2,4,7,3,5,6,8],[4,7,2,1,5,3,8,6]
复制
返回值：
{1,2,3,4,#,5,6,#,7,#,#,8}
复制
说明：
返回根节点，系统会输出整颗二叉树对比结果，重建结果如题面图示
