剑指offer | 二叉搜索树的后序遍历序列 | 29

目录

• 思路分析

思路分析

cpp

class Solution {
public:
    vector<int> nums;
    bool recur(int l,int r){
        if(l>=r)return true;
        int i = l;
        while(nums[i]<nums[r])i++;
        int j = i;
        while(nums[i]>nums[r])i++;
        return i==r && recur(l,j-1) && recur(j,r-1);
    }

    bool verifyPostorder(vector<int>& postorder) {
        nums = postorder;
        return recur(0,nums.size()-1);
    }
};

python

class Solution:
    def verifyPostorder(self, postorder: List[int]) -> bool:
        nums = postorder

        def recur(l:int,r:int)->bool:
            if l>=r:return True
            i = l
            while nums[i]<nums[r]:i+=1
            j = i
            while nums[i]>nums[r]:i+=1
            return i==r and recur(l,j-1) and recur(j,r-1)
        
        return recur(0,len(nums)-1)