998. Maximum Binary Tree II

We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.

Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with the following Construct(A) routine:

  • If A is empty, return null.
  • Otherwise, let A[i] be the largest element of A.  Create a root node with value A[i].
  • The left child of root will be Construct([A[0], A[1], ..., A[i-1]])
  • The right child of root will be Construct([A[i+1], A[i+2], ..., A[A.length - 1]])
  • Return root.

Note that we were not given A directly, only a root node root = Construct(A).

Suppose B is a copy of A with the value val appended to it.  It is guaranteed that B has unique values.

Return Construct(B).

 

Example 1:

Input: root = [4,1,3,null,null,2], val = 5
Output: [5,4,null,1,3,null,null,2]
Explanation: A = [1,4,2,3], B = [1,4,2,3,5]

Example 2:

Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]
Explanation: A = [2,1,5,4], B = [2,1,5,4,3]

Example 3:

Input: root = [5,2,3,null,1], val = 4
Output: [5,2,4,null,1,3]
Explanation: A = [2,1,5,3], B = [2,1,5,3,4]

 

Constraints:

  • 1 <= B.length <= 100

Rust Solution

struct Solution;
use rustgym_util::*;

trait Postorder {
    fn insert(self, val: i32) -> Self;
}

impl Postorder for TreeLink {
    fn insert(self, val: i32) -> Self {
        if let Some(node) = self {
            let node_val = node.borrow().val;
            if node_val < val {
                tree!(val, Some(node), None)
            } else {
                let right = node.borrow_mut().right.take();
                node.borrow_mut().right = right.insert(val);
                Some(node)
            }
        } else {
            tree!(val)
        }
    }
}

impl Solution {
    fn insert_into_max_tree(root: TreeLink, val: i32) -> TreeLink {
        root.insert(val)
    }
}

#[test]
fn test() {
    let root = tree!(4, tree!(1), tree!(3, tree!(2), None));
    let val = 5;
    let res = tree!(5, tree!(4, tree!(1), tree!(3, tree!(2), None)), None);
    assert_eq!(Solution::insert_into_max_tree(root, val), res);
    let root = tree!(5, tree!(2, None, tree!(1)), tree!(4));
    let val = 3;
    let res = tree!(5, tree!(2, None, tree!(1)), tree!(4, None, tree!(3)));
    assert_eq!(Solution::insert_into_max_tree(root, val), res);
    let root = tree!(5, tree!(2, None, tree!(1)), tree!(3));
    let val = 4;
    let res = tree!(5, tree!(2, None, tree!(1)), tree!(4, tree!(3), None));
    assert_eq!(Solution::insert_into_max_tree(root, val), res);
}

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