LeetCode #919 — MEDIUM

Complete Binary Tree Inserter

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

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.

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree · Design

Example 1

["CBTInserter","insert","insert","get_root"]
[[[1,2]],[3],[4],[]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #919: Complete Binary Tree Inserter
/**
 * 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 CBTInserter {
    private List<TreeNode> tree = new ArrayList<>();

    public CBTInserter(TreeNode root) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            for (int i = q.size(); i > 0; --i) {
                TreeNode node = q.poll();
                tree.add(node);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
    }

    public int insert(int val) {
        TreeNode p = tree.get((tree.size() - 1) / 2);
        TreeNode node = new TreeNode(val);
        tree.add(node);
        if (p.left == null) {
            p.left = node;
        } else {
            p.right = node;
        }
        return p.val;
    }

    public TreeNode get_root() {
        return tree.get(0);
    }
}

/**
 * Your CBTInserter object will be instantiated and called as such:
 * CBTInserter obj = new CBTInserter(root);
 * int param_1 = obj.insert(val);
 * TreeNode param_2 = obj.get_root();
 */

// Accepted solution for LeetCode #919: Complete Binary Tree Inserter
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type CBTInserter struct {
	tree []*TreeNode
}

func Constructor(root *TreeNode) CBTInserter {
	q := []*TreeNode{root}
	tree := []*TreeNode{}
	for len(q) > 0 {
		for i := len(q); i > 0; i-- {
			node := q[0]
			q = q[1:]
			tree = append(tree, node)
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	return CBTInserter{tree}
}

func (this *CBTInserter) Insert(val int) int {
	p := this.tree[(len(this.tree)-1)/2]
	node := &TreeNode{val, nil, nil}
	this.tree = append(this.tree, node)
	if p.Left == nil {
		p.Left = node
	} else {
		p.Right = node
	}
	return p.Val
}

func (this *CBTInserter) Get_root() *TreeNode {
	return this.tree[0]
}

/**
 * Your CBTInserter object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Insert(val);
 * param_2 := obj.Get_root();
 */

# Accepted solution for LeetCode #919: Complete Binary Tree Inserter
# 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 CBTInserter:

    def __init__(self, root: Optional[TreeNode]):
        self.tree = []
        q = deque([root])
        while q:
            for _ in range(len(q)):
                node = q.popleft()
                self.tree.append(node)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)

    def insert(self, val: int) -> int:
        p = self.tree[(len(self.tree) - 1) // 2]
        node = TreeNode(val)
        self.tree.append(node)
        if p.left is None:
            p.left = node
        else:
            p.right = node
        return p.val

    def get_root(self) -> Optional[TreeNode]:
        return self.tree[0]


# Your CBTInserter object will be instantiated and called as such:
# obj = CBTInserter(root)
# param_1 = obj.insert(val)
# param_2 = obj.get_root()

// Accepted solution for LeetCode #919: Complete Binary Tree Inserter
use rustgym_util::*;
struct CBTInserter {
    stack: Vec<TreeLink>,
}

impl CBTInserter {
    fn new(root: TreeLink) -> Self {
        let mut stack: Vec<TreeLink> = vec![root];
        let mut i = 0;
        while i < stack.len() {
            let left = stack[i].as_mut().unwrap().borrow_mut().left.clone();
            let right = stack[i].as_mut().unwrap().borrow_mut().right.clone();
            if left.is_some() {
                stack.push(left);
            }
            if right.is_some() {
                stack.push(right);
            }
            i += 1;
        }
        CBTInserter { stack }
    }

    fn insert(&mut self, v: i32) -> i32 {
        let link = tree!(v);
        let n = self.stack.len();
        self.stack.push(link.clone());
        let mut parent = self.stack[(n - 1) / 2].as_mut().unwrap().borrow_mut();
        if n % 2 == 1 {
            parent.left = link;
        } else {
            parent.right = link;
        }
        parent.val
    }

    fn get_root(&self) -> TreeLink {
        self.stack[0].clone()
    }
}

#[test]
fn test() {
    let mut obj = CBTInserter::new(tree!(1));
    assert_eq!(obj.insert(2), 1);
    assert_eq!(obj.get_root(), tree!(1, tree!(2), None));
}

// Accepted solution for LeetCode #919: Complete Binary Tree Inserter
/**
 * 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)
 *     }
 * }
 */

class CBTInserter {
    private tree: TreeNode[] = [];

    constructor(root: TreeNode | null) {
        if (root === null) {
            return;
        }
        const q: TreeNode[] = [root];
        while (q.length) {
            const t: TreeNode[] = [];
            for (const node of q) {
                this.tree.push(node);
                node.left !== null && t.push(node.left);
                node.right !== null && t.push(node.right);
            }
            q.splice(0, q.length, ...t);
        }
    }

    insert(val: number): number {
        const p = this.tree[(this.tree.length - 1) >> 1];
        const node = new TreeNode(val);
        this.tree.push(node);
        if (p.left === null) {
            p.left = node;
        } else {
            p.right = node;
        }
        return p.val;
    }

    get_root(): TreeNode | null {
        return this.tree[0];
    }
}

/**
 * Your CBTInserter object will be instantiated and called as such:
 * var obj = new CBTInserter(root)
 * var param_1 = obj.insert(val)
 * var param_2 = obj.get_root()
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.