Counting Maximal Value Roots in Binary Tree
Given a binary tree root, count and return the number of nodes where its value is greater than or equal to the values of all of its descendants.
For example, given
6
/ \
3 2
/ \
6 4Return 4 since all nodes except for 2 meet the criteria.
Example 1
Input
root = [6, [3, null, null], [2, [6, null, null], [4, null, null]]]Output
4# class Tree:
# def __init__(self, val, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def solve(self, root):
count = 0
def dfs(root):
if root:
left = dfs(root.left)
right = dfs(root.right)
if root.val >= left and root.val >= right:
nonlocal count
count += 1
return root.val
return max(left, right)
else:
return 0
dfs(root)
return count
We use a post order DFS here. Post order means:
Left children
Visits right children
Visit current node
We use post-order traversal to visit subtrees recursively. By visiting each subtree, we solve the problem at that subtree and eventually solve the full problem.
We need to do this because of this part of the question:
all of its descendants.
Our algorithm will go down to these subtrees first:
And then these:
And finally this:
The answer increments by 1 whenever node.val == subtree_max.
TODO write preorder s postorder vs in order
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