题解 | 重建二叉树

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param preOrder int整型vector 
     * @param vinOrder int整型vector 
     * @return TreeNode类
     */
      TreeNode* buildTree(vector<int>& preOrder,int left1,int right1,vector<int>& inOrder,int left2,int right2)
    {
        if(left1>right1||left2>right2)
        {
            return nullptr;
        }
        int rootlocation=-1;
        for(int i=left2;i<=right2;i++)
        {
            if(inOrder[i]==preOrder[left1])
            {
                rootlocation=i;
                break;
            }
        }
       TreeNode* root = new TreeNode(preOrder[left1]);
       TreeNode* left = buildTree(preOrder,left1+1,right1-(right2-rootlocation),inOrder,left2,rootlocation-1);//左子树的前序、中序
       TreeNode* right = buildTree(preOrder,right1-(right2-rootlocation)+1,right1,inOrder,rootlocation+1,right2);//右子树的前序、中序
       root->left=left;
       root->right=right;
       return root;
        
    }
    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        // write code here
        if(preOrder.size()==0||vinOrder.size()==0)
        {
            return nullptr;
        }
        return buildTree(preOrder,0,preOrder.size()-1,vinOrder,0,vinOrder.size()-1);
    }
};