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);
}
result.add(root.val);
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();
res.add(curr.val);
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();
result.add(node.val); // Add after all left children
p = node.right;
}
}
return result;
}