[LeetCode] 513. Find Bottom Left Tree Value

Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:

    2
   / \
  1   3

Output:
1
Example 2: 
Input:

        1
       / \
      2   3
     /   / \
    4   5   6
       /
      7

Output:
7
Note: You may assume the tree (i.e., the given root node) is not NULL.

Thought process:
BFS. Level order traversal. Use a variable to keep track of the first node of each level.

Solution 1 (BFS):
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int findBottomLeftValue(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        TreeNode bl = null;
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            bl = queue.peek();
            
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }
        return bl.val;
    }
}

Solution 2 (DFS):

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int deepest;
    private int left;
    
    public int findBottomLeftValue(TreeNode root) {
        left = root.val;
        findBottomLeftValue(root, 0);
        
        return left;
    }
    
    private void findBottomLeftValue(TreeNode root, int depth) {
        if (root == null) {
            return;
        }
        
        findBottomLeftValue(root.left, depth + 1);
        findBottomLeftValue(root.right, depth + 1);
        
        if (depth > deepest) {
            deepest = depth;
            left = root.val;
        }
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

Comments

Post a Comment

Popular posts from this blog

[LeetCode] 269. Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of  non-empty  words from the dictionary, where  words are sorted lexicographically by the rules of this new language . Derive the order of letters in this language. Example 1: Given the following words in dictionary, [ "wrt", "wrf", "er", "ett", "rftt" ] The correct order is:  "wertf" . Example 2: Given the following words in dictionary, [ "z", "x" ] The correct order is:  "zx" . Example 3: Given the following words in dictionary, [ "z", "x", "z" ] The order is invalid, so return  "" . Note: You may assume all letters are in lowercase. You may assume that if a is a prefix of b, then a must appear before b in the given dictionary. If the order is invalid, return an empty string
Read more

[LeetCode] 631. Design Excel Sum Formula

Your task is to design the basic function of Excel and implement the function of sum formula. Specifically, you need to implement the following functions: Excel(int H, char W):  This is the constructor. The inputs represents the height and width of the Excel form.  H  is a positive integer, range from 1 to 26. It represents the height.  W  is a character range from 'A' to 'Z'. It represents that the width is the number of characters from 'A' to  W . The Excel form content is represented by a height * width 2D integer array  C , it should be initialized to zero. You should assume that the first row of  C  starts from 1, and the first column of  C  starts from 'A'. void Set(int row, char column, int val):  Change the value at  C(row, column)  to be val. int Get(int row, char column):  Return the value at  C(row, column) . int Sum(int row, char column, List of Strings : numbers):  This function calculate and set the value at  C(row, column) , wh
Read more

[LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee

Your are given an array of integers  prices , for which the  i -th element is the price of a given stock on day  i ; and a non-negative integer  fee  representing a transaction fee. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.) Return the maximum profit you can make. Example 1: Input: prices = [1, 3, 2, 8, 4, 9], fee = 2 Output: 8 Explanation: The maximum profit can be achieved by: Buying at prices[0] = 1 Selling at prices[3] = 8 Buying at prices[4] = 4 Selling at prices[5] = 9 The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8. Note: 0 < prices.length <= 50000 . 0 < prices[i] < 50000 . 0 <= fee < 50000 . Thought process: Each day there are two options, buy or sell. Use two variables buy and sell to keep track of the max profit. Solution: 1 2 3 4 5 6
Read more