**Ways to Remove Leaves of a Tree - Google Top Interview Questions**

### Problem Statement :

You are given a binary tree root. In one operation you can delete one leaf node in the tree. Return the number of different ways there are to delete the whole tree. Constraints 0 ≤ n ≤ 100,000 Example 1 Input Visualize root = [1, [2, null, null], [3, [4, null, null], null]] Output 3 Explanation The three ways to delete are [2, 4, 3, 1] [4, 2, 3, 1] [4, 3, 2, 1] Solved 92 Attempted 127

### Solution :

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
static long[][] rd = new long[100001][101];
static final int MODE = 1000000007;
static {
for (int i = 0; i < rd.length; ++i) {
rd[i][0] = 1;
for (int j = 1; j < i && j < rd[0].length; ++j) {
rd[i][j] = (rd[i - 1][j - 1] + rd[i - 1][j]) % MODE;
}
if (i < rd[0].length)
rd[i][i] = 1;
}
}
public int solve(Tree root) {
int[] res = todo(root);
return res[0];
}
private int[] todo(Tree root) {
if (root == null)
return new int[] {1, 0};
int[] l = todo(root.left), r = todo(root.right);
long t;
if (l[1] <= r[1]) {
t = rd[l[1] + r[1]][l[1]];
} else {
t = rd[l[1] + r[1]][r[1]];
}
t *= l[0] * r[0];
l[0] = (int) (t % MODE);
l[1] += r[1] + 1;
return l;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, root):
def rec(root): # return (size, number of ways) for current subtree
if root is None: # Root is null
return (0, 1)
leftsize, leftdp = rec(root.left)
rightsize, rightdp = rec(root.right)
return (
leftsize + rightsize + 1,
comb(leftsize + rightsize, leftsize) * leftdp * rightdp,
)
return rec(root)[1]
themast3r
95
Big N
6 months ago
Came up with an exact solution lol. Was thinking it's a elegant solution, so, I should post XD
class Solution:
def solve(self, root):
def helper(root):
if not root:
return [0, 1]
leftSize, leftWays, rightSize, rightWays = helper(root.left) + helper(root.right)
return [ leftSize + rightSize + 1, leftWays * rightWays * math.comb(leftSize + rightSize, leftSize) ]
return helper(root).pop()
```
```

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