problemmediumalgorithmsleetcode-919leetcode 919leetcode919

Complete Binary Tree Inserter

MediumUpdated: Jul 7, 2025
Practice on:
LeetCode

Problem

A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.

Design an algorithm to insert a new node to a complete binary tree keeping it complete after the insertion.

Implement the CBTInserter class:

  • CBTInserter(TreeNode root) Initializes the data structure with the root of the complete binary tree.
  • int insert(int v) Inserts a TreeNode into the tree with value Node.val == val so that the tree remains complete, and returns the value of the parent of the inserted TreeNode.
  • TreeNode get_root() Returns the root node of the tree.

Examples

Example 1

**Input**
["CBTInserter", "insert", "insert", "get_root"]
[[[1, 2]], [3], [4], []]
**Output**
[null, 1, 2, [1, 2, 3, 4]]

**Explanation**
CBTInserter cBTInserter = new CBTInserter([1, 2]);
cBTInserter.insert(3);  // return 1
cBTInserter.insert(4);  // return 2
cBTInserter.get_root(); // return [1, 2, 3, 4]

Constraints

  • The number of nodes in the tree will be in the range [1, 1000].
  • 0 <= Node.val <= 5000
  • root is a complete binary tree.
  • 0 <= val <= 5000
  • At most 104 calls will be made to insert and get_root.

Solution

Method 1 – BFS Queue for Insertion Tracking

Intuition

To efficiently insert into a complete binary tree, we can keep track of nodes that are not full (i.e., have less than two children). By using a queue, we always know where the next insertion should happen to maintain completeness.

Approach

  1. Use a queue to store nodes that are potential parents for new nodes (nodes with less than two children).
  2. On initialization, perform a BFS traversal to fill the queue with such nodes.
  3. For insert(val), take the front node from the queue:
    • If it has no left child, insert as left child.
    • Else, insert as right child and remove the node from the queue (now full).
    • Add the new node to the queue (it may receive children in the future).
  4. For get_root(), return the root node.

Code

C++
class CBTInserter {
    TreeNode* r;
    std::queue<TreeNode*> q;
public:
    CBTInserter(TreeNode* root) : r(root) {
        std::queue<TreeNode*> t;
        t.push(root);
        while (!t.empty()) {
            auto n = t.front(); t.pop();
            if (!n->left || !n->right) q.push(n);
            if (n->left) t.push(n->left);
            if (n->right) t.push(n->right);
        }
    }
    int insert(int v) {
        auto n = q.front();
        TreeNode* node = new TreeNode(v);
        if (!n->left) n->left = node;
        else { n->right = node; q.pop(); }
        q.push(node);
        return n->val;
    }
    TreeNode* get_root() { return r; }
};
Go
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}
type CBTInserter struct {
    r *TreeNode
    q []*TreeNode
}
func Constructor(root *TreeNode) CBTInserter {
    t := []*TreeNode{root}
    q := []*TreeNode{}
    for len(t) > 0 {
        n := t[0]
        t = t[1:]
        if n.Left == nil || n.Right == nil {
            q = append(q, n)
        }
        if n.Left != nil { t = append(t, n.Left) }
        if n.Right != nil { t = append(t, n.Right) }
    }
    return CBTInserter{r: root, q: q}
}
func (this *CBTInserter) Insert(v int) int {
    n := this.q[0]
    node := &TreeNode{Val: v}
    if n.Left == nil {
        n.Left = node
    } else {
        n.Right = node
        this.q = this.q[1:]
    }
    this.q = append(this.q, node)
    return n.Val
}
func (this *CBTInserter) Get_root() *TreeNode {
    return this.r
}
Java
class CBTInserter {
    private TreeNode r;
    private Queue<TreeNode> q = new LinkedList<>();
    public CBTInserter(TreeNode root) {
        r = root;
        Queue<TreeNode> t = new LinkedList<>();
        t.offer(root);
        while (!t.isEmpty()) {
            TreeNode n = t.poll();
            if (n.left == null || n.right == null) q.offer(n);
            if (n.left != null) t.offer(n.left);
            if (n.right != null) t.offer(n.right);
        }
    }
    public int insert(int v) {
        TreeNode n = q.peek();
        TreeNode node = new TreeNode(v);
        if (n.left == null) n.left = node;
        else { n.right = node; q.poll(); }
        q.offer(node);
        return n.val;
    }
    public TreeNode get_root() { return r; }
}
Kotlin
class CBTInserter(root: TreeNode) {
    private val r = root
    private val q: ArrayDeque<TreeNode> = ArrayDeque()
    init {
        val t = ArrayDeque<TreeNode>()
        t.add(root)
        while (t.isNotEmpty()) {
            val n = t.removeFirst()
            if (n.left == null || n.right == null) q.add(n)
            n.left?.let { t.add(it) }
            n.right?.let { t.add(it) }
        }
    }
    fun insert(v: Int): Int {
        val n = q.first()
        val node = TreeNode(v)
        if (n.left == null) n.left = node
        else { n.right = node; q.removeFirst() }
        q.add(node)
        return n.`val`
    }
    fun get_root(): TreeNode = r
}
Python
class CBTInserter:
    def __init__(self, root: TreeNode):
        self.r = root
        self.q = []
        t = [root]
        while t:
            n = t.pop(0)
            if not n.left or not n.right:
                self.q.append(n)
            if n.left:
                t.append(n.left)
            if n.right:
                t.append(n.right)
    def insert(self, v: int) -> int:
        n = self.q[0]
        node = TreeNode(v)
        if not n.left:
            n.left = node
        else:
            n.right = node
            self.q.pop(0)
        self.q.append(node)
        return n.val
    def get_root(self) -> TreeNode:
        return self.r
Rust
use std::collections::VecDeque;
impl CBTInserter {
    pub fn new(root: TreeNode) -> Self {
        let mut t = VecDeque::new();
        let mut q = VecDeque::new();
        let r = Box::new(root);
        t.push_back(r.clone());
        while let Some(n) = t.pop_front() {
            if n.left.is_none() || n.right.is_none() {
                q.push_back(n.clone());
            }
            if let Some(ref l) = n.left { t.push_back(l.clone()); }
            if let Some(ref r) = n.right { t.push_back(r.clone()); }
        }
        CBTInserter { r, q }
    }
    pub fn insert(&mut self, v: i32) -> i32 {
        let n = self.q.front().unwrap().clone();
        let node = Box::new(TreeNode::new(v));
        if n.left.is_none() {
            n.left = Some(node);
        } else {
            n.right = Some(node);
            self.q.pop_front();
        }
        self.q.push_back(node);
        n.val
    }
    pub fn get_root(&self) -> Box<TreeNode> {
        self.r.clone()
    }
}
TypeScript
class CBTInserter {
    private r: TreeNode;
    private q: TreeNode[] = [];
    constructor(root: TreeNode) {
        this.r = root;
        const t: TreeNode[] = [root];
        while (t.length) {
            const n = t.shift()!;
            if (!n.left || !n.right) this.q.push(n);
            if (n.left) t.push(n.left);
            if (n.right) t.push(n.right);
        }
    }
    insert(v: number): number {
        const n = this.q[0];
        const node = new TreeNode(v);
        if (!n.left) n.left = node;
        else { n.right = node; this.q.shift(); }
        this.q.push(node);
        return n.val;
    }
    get_root(): TreeNode {
        return this.r;
    }
}

Complexity

  • ⏰ Time complexity: O(1) per insert
  • 🧺 Space complexity: O(n)

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