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) {}
 * };
 */
#include <vector>
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* rebuild(TreeNode* root,vector<int> preOrder,vector<int> vinOrder,int preL,int preR,int inL,int inR){
        if(preL>preR)
            return nullptr;

        root=new TreeNode(preOrder[preL]);
        int index;
        for(int i=inL;i<=inR;i++){
            if(vinOrder[i]==preOrder[preL]){
                index=i;
                break;
            }
        }
        int leftNum=index-inL;
        root->left=rebuild(root->left,preOrder,vinOrder,preL+1,preL+leftNum,inL,index-1);
        root->right=rebuild(root->right,preOrder,vinOrder,preL+leftNum+1,preR,index+1,inR);
        return root;
    }
    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) {
        // write code here
        TreeNode* root=rebuild(root,preOrder,vinOrder,0,preOrder.size()-1,0,vinOrder.size()-1);
        return root;

    }
};

根据推导结果，可以得到上面的递归表达