9. Binary Tree Inorder Traversal
easyAsked at GitLabReturn the in-order traversal of a binary tree's node values.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given the root of a binary tree, return an array of node values produced by in-order (left, node, right) traversal.
Constraints
0 <= nodes <= 100-100 <= val <= 100
Examples
Example 1
Input
root=[1,null,2,3]Output
[1,3,2]Example 2
Input
root=[]Output
[]Approaches
1. Recursive
Recurse left, push value, recurse right.
- Time
- O(n)
- Space
- O(h)
const out=[];
const dfs=n=>{ if(!n)return; dfs(n.left); out.push(n.val); dfs(n.right); };
dfs(root); return out;Tradeoff:
2. Iterative with stack
Push lefts, pop, visit, then move to right.
- Time
- O(n)
- Space
- O(h)
function inorder(root){
const out=[], st=[]; let n=root;
while (n || st.length){
while (n){ st.push(n); n=n.left; }
n=st.pop(); out.push(n.val); n=n.right;
}
return out;
}Tradeoff:
GitLab-specific tips
GitLab values explaining why the iterative version is preferable when traversing massive merge-request comment trees on a single runner.
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