# C++ O(n log n) recursive solution.

• I'm in bit of disbelief that my solution was accepted. I coded it up while I was really sleepy, and hit submit, fully expecting it to fail compilation or something...

The logic is as follows:

For any pre-order sequence, we have:

1. The root, followed by
2. Another preorder sequence for the LHS, where everything is less than root, followed by,
3. Another preorder sequence for the RHS, where everything is greater than root.

In order to find the boundary of the LHS preorder and the RHS preorder, you can run binary search to look for the boundary where the elements transition from being (< root) to being (> root).

Once you find the boundary, you can recursively call the verify function on the LHS preorder sequence and the RHS preorder sequence. The trick is to pass two elements along, that say what the relationship of the elements should be w.r.t. the ancestors that have gone by so far. e.g. I am passing two elements (lessThan and greaterThan) which say that all elements in the preorder sequence need to be less than lessThan, and greater than greaterThan. Look at the recursive calls for lhsverify and rhsverify bools to see what is being passed on to recursive calls.

``````class Solution {
//boundary denotes LAST number that is less than or equal to num
int binarySearchBoundary(vector<int>& preorder, int from, int to, int num) {
if (from == to) return from;
int mid = (from+to)/2;
int midplus = mid + 1;
if (preorder[mid] <= num && preorder[midplus] > num) return mid;
if (preorder[mid] > num) return binarySearchBoundary(preorder, from, mid, num);
return binarySearchBoundary(preorder, midplus, to, num);
}

//verify the sequence from "from" to "to", in which every elements needs to be
// less than "lessThan", and greater than "greaterThan"
bool verify(vector<int>& preorder, int from, int to, int lessThan, int greaterThan) {
if (from > to) return true;
int root = preorder[from];
if (root > lessThan || root < greaterThan) return false;
if (from == to) return true;

int boundary = binarySearchBoundary(preorder, from, to, root);
bool lhsverify = verify(preorder, from+1, boundary, min(lessThan, root), greaterThan);
bool rhsverify = verify(preorder, boundary+1, to, lessThan, max(root, greaterThan));
return (lhsverify && rhsverify);
}

public:
bool verifyPreorder(vector<int>& preorder) {
if (preorder.size() == 0) return true;
return verify(preorder, 0, preorder.size()-1, std::numeric_limits<int>::max(), std::numeric_limits<int>::min());
}
};``````

• Mine is similar, just as a reference :)

``````class Solution {
bool verifyPreorder(vector<int>& preorder, int begin, int end, int &larger_than) {
int n = end - begin;
// if n <= 2, it can be a valid subtree as long as its values are smaller than its root
if (n <= 2)   {
for (int i = begin; i < end; i++)   {
if (preorder[i] < larger_than)  {
return false;
}
}
return true;
}

int val_root = preorder[begin];
if (val_root < larger_than) {
return false;
}

int end_left = begin + 1;
if (preorder[end_left] < val_root)   {
// has left child
while (end_left < end && preorder[end_left] < val_root)   {
end_left++;
}
}

bool left = verifyPreorder(preorder, begin + 1, end_left, larger_than);
if (left)   {
// if left is ok, check right
// right's value should be larger than root's value
larger_than = val_root;
return verifyPreorder(preorder, end_left, end, larger_than);
} else  {
return false;
}
}

public:
bool verifyPreorder(vector<int>& preorder) {
int larger_than = INT_MIN;
return verifyPreorder(preorder, 0, preorder.size(), larger_than);
}
};
``````

• Amazing algorithm, unexpected but beautiful.

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