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Program to find number of good leaf nodes pairs using Python
Suppose we have a binary tree. and another value distance d. A pair of two different leaf nodes are said to be good, when the shortest path between these two nodes is smaller or same as distance d.
So, if the input is like
And distance d = 4, then the output will be 2 because the pairs are (8,7) and (5,6) as their path length distance is 2, but (7,5) or (8,6) or other pairs are not good as their path length is 5 which is larger than d = 4
To solve this, we will follow these steps −
sol := 0
Define a function util() . This will take root
if root is null, then
return a new list
if root is leaf, then
return an array with one pair [0, 0]
otherwise,
cur := a new list
l := util(left of root)
r := util(right of root)
for each n in l, do
n[1] := n[1] + 1
for each n in r, do
n[1] := n[1] + 1
for each n in r, do
for each n1 in l, do
if n[1] + n1[1] <= d, then
sol := sol + 1
return l+r
From the main method do the following −
util(root)
return sol
Let us see the following implementation to get better understanding −
Example
class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: def __init__(self): self.sol = 0 def solve(self, root, d): def util(root): if not root: return [] if not root.left and not root.right: return [[0, 0]] else: cur = [] l = util(root.left) r = util(root.right) for n in l: n[1] += 1 for n in r: n[1] += 1 for n in r: for n1 in l: if n[1] + n1[1] <= d: self.sol += 1 return l+r util(root) return self.sol root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) root.left.right = TreeNode(4) root.left.right.left = TreeNode(8) root.left.right.right = TreeNode(7) root.right.left = TreeNode(5) root.right.right = TreeNode(6) d = 4 ob = Solution() print(ob.solve(root, d))
Input
root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) root.left.right = TreeNode(4) root.left.right.left = TreeNode(8) root.left.right.right = TreeNode(7) root.right.left = TreeNode(5) root.right.right = TreeNode(6) d = 4
Output
2
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