面试题33

输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历结果。如果是则返回 true，否则返回 false。假设输入的数组的任意两个数字都互不相同。

参考以下这颗二叉搜索树：

     5
    / \
   2   6
  / \
 1   3

示例 1：

输入: [1,6,3,2,5]
输出: false

示例 2：

输入: [1,3,2,6,5]
输出: true

提示：

• 数组长度 <= 1000

Solutions

1. mono stack

2. reference: https://leetcode-cn.com/problems/er-cha-sou-suo-shu-de-hou-xu-bian-li-xu-lie-lcof/solution/dan-diao-di-zeng-zhan-by-shi-huo-de-xia-tian/
3. This solution is based on the fact that postorder traversal is the opposite of preorder traversal. ie: preorder(Similar): root right left -> postorder: left right root

4. Maintain a monotonically increasing stack which records the path when traversing the right subtree(Mono increasing).

5. This solution works only if there are no duplicate elements in the given sequence.

class Solution {
public:
    bool verifyPostorder(vector<int>& postorder) {
        stack<int> s;

        int cur = postorder.size() - 1;
        int rootval = INT_MAX;

        for (int cur = postorder.size() - 1; cur >= 0; cur--) {

            if (postorder[cur] > rootval)
                return false;
            // find the root node
            while (!s.empty() && s.top() > postorder[cur]) {
                rootval = s.top();
                s.pop();
            }
            s.push(postorder[cur]);
        }

        return true;
    }
};

1. recursion

2. In each recursive call, find the first element of the right subtree(val > root->val).

class Solution {
public:
    bool valid(vector<int> & postorder, int lo, int hi) {
        if (lo >= hi)
            return true;
        // the root val is the last element
        int rootval = postorder[hi];
        int mid = lo;
        while (postorder[mid] < rootval && mid < hi)
            mid++;
        // check if the right tree is valid
        for (int r = mid; r < hi; r++)
            if (postorder[r] < rootval)
                return false;

        return valid(postorder, lo, mid - 1) && valid(postorder, mid, hi - 1);
    }

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