https://github.com/emahtab/binary-tree-inorder-traversal
Binary Tree Inorder Traversal
https://github.com/emahtab/binary-tree-inorder-traversal
binary-tree inorder-traversal leetcode problem-solving
Last synced: 3 months ago
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Binary Tree Inorder Traversal
- Host: GitHub
- URL: https://github.com/emahtab/binary-tree-inorder-traversal
- Owner: eMahtab
- Created: 2020-01-12T04:51:29.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2020-02-06T08:38:08.000Z (over 5 years ago)
- Last Synced: 2025-02-02T03:26:17.086Z (5 months ago)
- Topics: binary-tree, inorder-traversal, leetcode, problem-solving
- Homepage:
- Size: 53.7 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Binary Tree Inorder Traversal 🌲
## https://leetcode.com/problems/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?**
# Inorder Traversal :
Note that node 75 doesn't have left child and node 29 doesn't have right child.
The inorder traversal for above binary tree will be **[21, 35, 20, 67, 75, 30, 70, 50, 29, 43, 60, 24, 65]**
## Implementation : Recursive
```java
import java.util.ArrayList;
import java.util.List;
public class App {
public static void main(String[] args) {
TreeNode root = new TreeNode(70);
root.left = new TreeNode(67); root.right = new TreeNode(43);
root.left.left = new TreeNode(35); root.left.right = new TreeNode(75);
root.left.left.left = new TreeNode(21); root.left.left.right = new TreeNode(20);
root.left.right.right = new TreeNode(30);
root.right.left = new TreeNode(29); root.right.left.left = new TreeNode(50);
root.right.right = new TreeNode(24);
root.right.right.left = new TreeNode(60); root.right.right.right = new TreeNode(65);
System.out.println(inorderTraversal(root));
}
// Definition for a binary tree node.
static public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public static List inorderTraversal(TreeNode root) {
List res = new ArrayList();
helper(root, res);
return res;
}
public static void helper(TreeNode root, List res) {
if (root != null) {
helper(root.left, res);
res.add(root.val);
helper(root.right, res);
}
}
}
```
## Implementation : Iterative (Simple and awesome ...)
```java
public List inorderTraversal(TreeNode root) {
List res = new ArrayList();
Stack stack = new Stack();
TreeNode curr = root;
// if current node is null and stack is also empty, we're done
while (!stack.empty() || curr != null) {
// if current node is not null, push it to the stack (defer it)and move to its left child
if (curr != null) {
stack.push(curr);
curr = curr.left;
} else {
// else if current node is null, we pop an element from stack,
// print it and finally set current node to its right child
curr = stack.pop();
res.add(curr.val);
curr = curr.right;
}
}
return res;
}
```
# References :
1. https://leetcode.com/problems/binary-tree-inorder-traversal/solution
2. https://www.youtube.com/watch?v=50v1sJkjxoc (Iterative Approach using Stack)
3. https://www.techiedelight.com/inorder-tree-traversal-iterative-recursive