513. Find Bottom Left Tree Value LeetCode Solution
In this guide, you will get 513. Find Bottom Left Tree Value LeetCode Solution with the best time and space complexity. The solution to Find Bottom Left Tree Value problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Table of Contents
- Problem Statement
- Complexity Analysis
- Find Bottom Left Tree Value solution in C++
- Find Bottom Left Tree Value solution in Java
- Find Bottom Left Tree Value solution in Python
- Additional Resources
Problem Statement of Find Bottom Left Tree Value
Given the root of a binary tree, return the leftmost value in the last row of the tree.
Example 1:
Input: root = [2,1,3]
Output: 1
Example 2:
Input: root = [1,2,3,4,null,5,6,null,null,7]
Output: 7
Constraints:
The number of nodes in the tree is in the range [1, 104].
-231 <= Node.val <= 231 – 1
Complexity Analysis
- Time Complexity: O(n)
- Space Complexity: O(n)
513. Find Bottom Left Tree Value LeetCode Solution in C++
class Solution {
public:
int findBottomLeftValue(TreeNode* root) {
queue<TreeNode*> q{{root}};
TreeNode* node = nullptr;
while (!q.empty()) {
node = q.front();
q.pop();
if (node->right)
q.push(node->right);
if (node->left)
q.push(node->left);
}
return node->val;
}
};
/* code provided by PROGIEZ */
513. Find Bottom Left Tree Value LeetCode Solution in Java
class Solution {
public int findBottomLeftValue(TreeNode root) {
Queue<TreeNode> q = new ArrayDeque<>(List.of(root));
TreeNode node = null;
while (!q.isEmpty()) {
node = q.poll();
if (node.right != null)
q.offer(node.right);
if (node.left != null)
q.offer(node.left);
}
return node.val;
}
}
// code provided by PROGIEZ
513. Find Bottom Left Tree Value LeetCode Solution in Python
class Solution:
def findBottomLeftValue(self, root: TreeNode | None) -> int:
q = collections.deque([root])
while q:
root = q.popleft()
if root.right:
q.append(root.right)
if root.left:
q.append(root.left)
return root.val
# code by PROGIEZ
Additional Resources
- Explore all LeetCode problem solutions at Progiez here
- Explore all problems on LeetCode website here
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