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
SOLUTION:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
nodes = [(root, 0, 0)]
leftmost = (0, 0, root.val)
while len(nodes) > 0:
curr, w, l = nodes.pop()
if curr:
if l > leftmost[1] or (l == leftmost[1] and w < leftmost[0]):
leftmost = (w, l, curr.val)
nodes.append((curr.left, w - 1, l + 1))
nodes.append((curr.right, w + 1, l + 1))
return leftmost[2]
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