# Python solution with detailed explanation

• Second Minimum Node In a Binary Tree https://leetcode.com/problems/second-minimum-node-in-a-binary-tree/description/

Using BFS for second minimum

• The root is the minimum value.
• Beginning from second level, the smallest value from the root is a candidate for second minimum.
• Now if there is no value on a level which is same as the root, then the second minimum value is the minimum value of that level. Otherwise, the minimum level is just a potential candidate. We need to continue searching until we find a level which has no value different from the root, or we exhaust all the levels.
``````from collections import deque
class Solution:
def findSecondMinimumValue(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return -1
dq = deque()
second_val = float('inf')
if root.left:
dq.append(root.left)
dq.append(root.right)
while dq:
root_value_found = False
for _ in range(len(dq)):
x = dq.popleft()
if root.val < x.val < second_val:
second_val = x.val
elif x.val == root.val:
root_value_found = True
if x.left:
dq.append(x.left)
dq.append(x.right)
if second_val != float('inf') and not root_value_found:
return second_val
return second_val if second_val != float('inf') else -1
``````

DFS Based Search

``````class Solution:
def findSecondMinimumValue(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return -1
second_min = float('inf')
def helper(x, min_val, t):
if min_val < x.val < t[0]:
t[0] = x.val
if x.left:
helper(x.left, min_val, t)
helper(x.right, min_val, t)
return
t = [second_min]
helper(root, root.val, t)
second_min = t[0]
return second_min if second_min != float('inf') else -1
``````

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