Leetcode 144 & 94 & 145 Binary Tree Preorder/Inorder/Postorder Traversal

Given a binary tree, return the preorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

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

Time complexity: O(n)

Space complexity: O(n)

Code is below:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List preorderTraversal(TreeNode root) {
        // O(n) O(n)
        
        // recursive
        // LinkedList result = new LinkedList<>();
        // traverse(root, result);
        // return result;
        
        // non-recursive
        LinkedList result = new LinkedList<>();
        Stack stack = new Stack<>();
        
        if (root == null) {
            return result;
        }
        
        stack.push(root);
        
        while(!stack.empty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            if (node.right != null) {
                stack.push(node.right);
            }
            
            if (node.left != null) {
                stack.push(node.left);
            }
        }
        
        return result;
    }
    
    public void traverse(TreeNode root, LinkedList result) {
        if (root == null) {
            return;
        }
        
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}

Inorder:
Given a binary tree, return the inorder traversal of its nodes’ values.

Code is below:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List inorderTraversal(TreeNode root) {
        // O(n) O(n)
        
        // LinkedList result = new LinkedList<>();
        // traverse(root, result);
        // return result;
        
        LinkedList result = new LinkedList<>();
        Stack stack = new Stack();
        
        if (root == null) {
            return result;
        }
        
        TreeNode cur = root;
        
        while (!stack.empty() || cur != null) {
            while (cur != null) {
                stack.add(cur);
                cur = cur.left;
            }
            
            cur = stack.peek();
            stack.pop();
            result.add(cur.val);
            cur = cur.right;
        }
        
        return result;
    }
    
    public void traverse(TreeNode root, LinkedList result) {
        if (root == null) {
            return;
        }
        
        traverse(root.left, result);
        result.add(root.val);
        traverse(root.right, result);
    }
}

Postorder:
Given a binary tree, return the postorder traversal of its nodes’ values.

Code is below:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List postorderTraversal(TreeNode root) {
        // O(n) O(n)
        
        LinkedList result = new LinkedList<>();
        traverse(root, result);
        
        return result;
    }
    
    public void traverse(TreeNode root, LinkedList result) {
        if (root == null) {
            return;
        }
        
        traverse(root.left, result);
        traverse(root.right, result);
        result.add(root.val);
    }
}