# Binary Tree Inorder Traversal

Given a binary tree, return the_inorder_traversal of its nodes' values.

Example:

``````Input:
[1,null,2,3]
1
\
2
/
3

Output:
[1,3,2]``````

Follow up:Recursive solution is trivial, could you do it iteratively?

## Analysis

Approach 1 - Recursive

Following the definition of inorder traversal: left - root - right

## Solution

### Recursive

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
recursiveInorderTravesal(root, result);
return result;
}
void recursiveInorderTravesal(TreeNode root, List<Integer> result) {
if (root != null) {
if (root.left != null) {
recursiveInorderTravesal(root.left, result);
}
if (root.right != null) {
recursiveInorderTravesal(root.right, result);
}
}
}
}``````

TIme complexity: O(n)

Space Complexity: worst case O(n), average O(log(n))

### Iterative (Using Stack)

LeetCode Official Solution In-order Traversal

``````public class Solution {
public List < Integer > inorderTraversal(TreeNode root) {
List < Integer > res = new ArrayList < > ();
Stack < TreeNode > stack = new Stack < > ();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
curr = stack.pop();
curr = curr.right;
}
return res;
}
}``````

Another Stack Implementation

``````public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode p = root;
while(!stack.isEmpty() || p != null) {
if(p != null) {
stack.push(p);
p = p.left;
} else {
TreeNode node = stack.pop();