Tree

Binary Tree Inorder Traversal

Link: https://leetcode.com/problems/binary-tree-inorder-traversal/

Code:

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

• inorder traversal: left, root, right;

• preorder traversal: root, left, right;

• postorder traversal: left, right, root;

• add() method for List;

Unique Binary Search Trees

Link: https://leetcode.com/problems/unique-binary-search-trees/

Code:

class Solution {
    public int numTrees(int n) {
        int[] G = new int[n + 1];
        G[0] = 1;
        G[1] = 1;
        for (int i = 2; i <= n; i++) {
            for (int j = 1; j <= i; j++) {
                G[i] += G[j-1] * G[i-j];
            }
        }
        return G[n];
    }
}

• Dynamic programming;

• G[i] is the number of unique BSTs with i nodes;

• Base case: G[0] = 1, G[1] = 1;

• Recursive case: G[i] = G[0] * G[i-1] + G[1] * G[i-2] + ... + G[i-1] * G[0];