10. Binary Tree Inorder Traversal
easyAsked at CourseraReturn inorder traversal of a binary tree — Coursera tests recursive-vs-iterative tradeoffs in a course-tree walking context.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given the root of a binary tree, return the inorder (left, node, right) traversal of its nodes' values.
Constraints
0 <= nodes <= 100-100 <= Node.val <= 100
Examples
Example 1
root = [1,null,2,3][1,3,2]Example 2
root = [][]Approaches
1. Recursion
Walk left, push node, walk right.
- Time
- O(n)
- Space
- O(h)
function inorder(root, out = []) {
if (!root) return out;
inorder(root.left, out);
out.push(root.val);
inorder(root.right, out);
return out;
}Tradeoff:
2. Iterative stack
Push left chain, pop and visit, descend right.
- Time
- O(n)
- Space
- O(h)
function inorder(root) {
const out = [], stack = [];
let cur = root;
while (cur || stack.length) {
while (cur) { stack.push(cur); cur = cur.left; }
cur = stack.pop();
out.push(cur.val);
cur = cur.right;
}
return out;
}Tradeoff:
Coursera-specific tips
Coursera reviewers value showing the iterative variant even after the recursive one — they ship traversal in browser environments where stack depth has bitten them before.
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