Leetcode 270. Closest Binary Search Tree Value

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

Note:

• Given target value is a floating point.

• You are guaranteed to have only one unique value in the BST that is closest to the target.

Analysis:

Maintain a min value and update it with left and right .

Time complexity: O(logn)

Space complexity: O(1)

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 int closestValue(TreeNode root, double target) {
        // O(logn)
        
        int min = root.val;
        
        while (root != null) {
            min = Math.abs(target - root.val) < Math.abs(target - min) ? root.val : min;
            root = root.val < target ? root.right : root.left;
        }
        
        return min;
    }
}

Leetcode 102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Analysis:

Use queue.

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<List> levelOrder(TreeNode root) {
        // O(n) BFS
        
        List<List> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        Queue queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            ArrayList level = new ArrayList<>();
            int size = queue.size();
            
            for (int i = 0; i < size; i++) {
                // poll the head
                TreeNode head = queue.poll();
                level.add(head.val);
                
                // add the left subnode
                if (head.left != null) {
                    queue.offer(head.left);
                }
                
                // add the right subnode
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            result.add(level);
        }
        return result;
    }
}

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);
    }
}

Leetcode 235 & 236. Lowest Common Ancestor of a Binary (Search) Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Analysis:

Check for the properties of the BST. If root.val > both p.val and q.val, we know the lca is in the left side, then goto the left subtree. If root.val < both p.val & q.val, then go to right subtree. Otherwise, return root.

Time complexity: O(logn)

Space complexity: O(1)

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 TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // O(logn)
        
        if (root == null || p == null || q == null) {
            return null;
        }
        
        if (root.val > p.val && root.val > q.val) {
            // left sub tree
            return lowestCommonAncestor(root.left, p, q);
        } else if (root.val < p.val && root.val < q.val) {
            // right sub tree
            return lowestCommonAncestor(root.right, p, q);
        } else {
            return root;
        }
    }
}

Followup:

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Analysis:

No properties similar to BST. We can use divide and conquer. Divide the tree to be the left and right. If left & right not null, return root. If left not null, return left, otherwise return right.

Time complexity: O(n)

Space complexity: O(1)

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 TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        // O(n)
        
        if (root == null || root == p || root == q) {
            return root;
        }
        
        // divide: get the left and right lca
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        
        // conquer: check for the left and right
        // If left and right both exits, then return root
        if (left != null && right != null) {
            return root;
        }
        
        if (left != null) {
            return left;
        }
        
        if (right != null) {
            return right;
        }
        
        return null;
    }
}