LeetCode-面试题33-二叉搜索树的后序遍历序列

     5
    / \
   2   6
  / \
 1   3

输入: [1,6,3,2,5]
输出: false

输入: [1,3,2,6,5]
输出: true

class Solution {
    public boolean verifyPostorder(int[] postorder) {
        return helper(postorder,0,postorder.length-1);
    }

    public boolean helper(int[] postorder,int start,int end){
        if(start>=end) return true;
        int i = 0;
        for(i=start;i<end;i++){
            if(postorder[i]>postorder[end])
                break;
        }
        for(int j = i;j<end;j++){
            if(postorder[j]<postorder[end])
                return false;
        }
        return helper(postorder,start,i-1)&&helper(postorder,i,end-1);
    }
}

class Solution:
    def verifyPostorder(self, postorder: List[int]) -> bool:
        length = len(postorder);
        if not postorder:
            return True
        root = postorder[-1]
        # 二叉树中左子树始终比根节点小，右子树比根节点大
        # 寻找左子树边界
        i = 0
        for i in range(length):
            if postorder[i]>root:
                break;
        j = i
        for j in range(j,length):
            if postorder[j]<root:
                return False
        # 判断左子树是不是二叉搜索树
        left = True
        if i>0: left = self.verifyPostorder(postorder[:i])
        # 判断右子树是不是二叉搜索树
        right = True
        if i<length-1: right = self.verifyPostorder(postorder[i:-1])
        return left and right