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.
Build confidence with an intuition-first walkthrough focused on tree fundamentals.
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3] Output: [2,3,1]
Example 3:
Input: root = [] Output: []
Constraints:
[0, 100].-100 <= Node.val <= 100Problem summary: Given the root of a binary tree, invert the tree, and return its root. Example 1: Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1] Example 2: Input: root = [2,1,3] Output: [2,3,1] Example 3: Input: root = [] Output: []
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[4,2,7,1,3,6,9]
[2,1,3]
[]
reverse-odd-levels-of-binary-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #226: Invert Binary Tree
/**
* 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 invertTree(TreeNode root) {
if (root == null) {
return null;
}
TreeNode l = invertTree(root.left);
TreeNode r = invertTree(root.right);
root.left = r;
root.right = l;
return root;
}
}
// Accepted solution for LeetCode #226: Invert Binary Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
if root == nil {
return root
}
l, r := invertTree(root.Left), invertTree(root.Right)
root.Left, root.Right = r, l
return root
}
# Accepted solution for LeetCode #226: Invert Binary Tree
# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if root is None:
return None
l, r = self.invertTree(root.left), self.invertTree(root.right)
root.left, root.right = r, l
return root
// Accepted solution for LeetCode #226: Invert Binary Tree
// 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 invert_tree(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
if let Some(node) = root.clone() {
let mut node = node.borrow_mut();
let left = node.left.take();
let right = node.right.take();
node.left = Self::invert_tree(right);
node.right = Self::invert_tree(left);
}
root
}
}
// Accepted solution for LeetCode #226: Invert Binary Tree
/**
* 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 invertTree(root: TreeNode | null): TreeNode | null {
if (!root) {
return root;
}
const l = invertTree(root.left);
const r = invertTree(root.right);
root.left = r;
root.right = l;
return root;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.