Skip to content

998. Maximum Binary Tree II

DifficultyMedium

Description

A maximum tree is a tree where every node has a value greater than any other value in its subtree.

You are given the root of a maximum binary tree and an integer val.

Just as in the previous problem, the given tree was constructed from a 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 the 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:

  • The number of nodes in the tree is in the range [1, 100].
  • 1 <= Node.val <= 100
  • All the values of the tree are unique.
  • 1 <= val <= 100

Solutions

Solution 1: Recursion

Thinking

Append \(\textit{val}\) to the array that built this maximum tree and rebuild. If \(\textit{val}\) exceeds the root it becomes the new root with the old tree on the left; otherwise \(\textit{val}\) can only sit on the right spine. Recurse down the right child.

If \(\textit{val}\) is the maximum number, then make \(\textit{val}\) the new root node, and \(\textit{root}\) the left subtree of the new root node.

If \(\textit{val}\) is not the maximum number, since \(\textit{val}\) is the last appended number, it must be on the right side of \(\textit{root}\). Therefore, we can insert \(\textit{val}\) as a new node into the right subtree of \(\textit{root}\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) is the number of nodes in the tree.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoMaxTree(
        self, root: Optional[TreeNode], val: int
    ) -> Optional[TreeNode]:
        if root is None or root.val < val:
            return TreeNode(val, root)
        root.right = self.insertIntoMaxTree(root.right, val)
        return root

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoMaxTree(TreeNode root, int val) {
        if (root == null || root.val < val) {
            return new TreeNode(val, root, null);
        }
        root.right = insertIntoMaxTree(root.right, val);
        return root;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
        if (!root || root->val < val) return new TreeNode(val, root, nullptr);
        root->right = insertIntoMaxTree(root->right, val);
        return root;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func insertIntoMaxTree(root *TreeNode, val int) *TreeNode {
    if root == nil || root.Val < val {
        return &TreeNode{val, root, nil}
    }
    root.Right = insertIntoMaxTree(root.Right, val)
    return root
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function insertIntoMaxTree(root: TreeNode | null, val: number): TreeNode | null {
    if (!root || root.val < val) {
        return new TreeNode(val, root);
    }
    root.right = insertIntoMaxTree(root.right, val);
    return root;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn insert_into_max_tree(
        mut root: Option<Rc<RefCell<TreeNode>>>,
        val: i32,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        if root.is_none() || root.as_ref().unwrap().as_ref().borrow().val < val {
            return Some(Rc::new(RefCell::new(TreeNode {
                val,
                left: root.take(),
                right: None,
            })));
        }
        {
            let mut root = root.as_ref().unwrap().as_ref().borrow_mut();
            root.right = Self::insert_into_max_tree(root.right.take(), val);
        }
        root
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* insertIntoMaxTree(struct TreeNode* root, int val) {
    if (!root || root->val < val) {
        struct TreeNode* res = (struct TreeNode*) malloc(sizeof(struct TreeNode));
        res->val = val;
        res->left = root;
        res->right = NULL;
        return res;
    }
    root->right = insertIntoMaxTree(root->right, val);
    return root;
}

Solution 2: Iteration

Thinking

Recursion may follow the full height. Walk the right chain iteratively, stop at the first right child smaller than \(\textit{val}\), and hang the old right subtree on the new node's left, avoiding the call stack.

Search the right subtree, find the node where \(\textit{curr}.\textit{val} \gt \textit{val} \gt \textit{curr}.\textit{right}.\textit{val}\), then create a new node \(\textit{node}\), point \(\textit{node}.\textit{left}\) to \(\textit{curr}.\textit{right}\), and then point \(\textit{curr}.\textit{right}\) to \(\textit{node}\).

Finally, return \(\textit{root}\).

The time complexity is \(O(n)\), and the space complexity is \(O(1)\), where \(n\) is the number of nodes in the tree.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoMaxTree(
        self, root: Optional[TreeNode], val: int
    ) -> Optional[TreeNode]:
        if root.val < val:
            return TreeNode(val, root)
        curr = root
        node = TreeNode(val)
        while curr.right and curr.right.val > val:
            curr = curr.right
        node.left = curr.right
        curr.right = node
        return root

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoMaxTree(TreeNode root, int val) {
        if (root.val < val) {
            return new TreeNode(val, root, null);
        }
        TreeNode curr = root;
        TreeNode node = new TreeNode(val);
        while (curr.right != null && curr.right.val > val) {
            curr = curr.right;
        }
        node.left = curr.right;
        curr.right = node;
        return root;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
        if (root->val < val) return new TreeNode(val, root, nullptr);
        TreeNode* curr = root;
        TreeNode* node = new TreeNode(val);
        while (curr->right && curr->right->val > val) curr = curr->right;
        node->left = curr->right;
        curr->right = node;
        return root;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func insertIntoMaxTree(root *TreeNode, val int) *TreeNode {
    if root.Val < val {
        return &TreeNode{val, root, nil}
    }
    node := &TreeNode{Val: val}
    curr := root
    for curr.Right != nil && curr.Right.Val > val {
        curr = curr.Right
    }
    node.Left = curr.Right
    curr.Right = node
    return root
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function insertIntoMaxTree(root: TreeNode | null, val: number): TreeNode | null {
    if (root.val < val) {
        return new TreeNode(val, root);
    }
    const node = new TreeNode(val);
    let curr = root;
    while (curr.right && curr.right.val > val) {
        curr = curr.right;
    }
    node.left = curr.right;
    curr.right = node;
    return root;
}

Comments