<p>Given the <code>root</code> of a binary tree, replace the value of each node in the tree with the <strong>sum of all its cousins&#39; values</strong>.</p>

<p>Two nodes of a binary tree are <strong>cousins</strong> if they have the same depth with different parents.</p>

<p>Return <em>the </em><code>root</code><em> of the modified tree</em>.</p>

<p><strong>Note</strong> that the depth of a node is the number of edges in the path from the root node to it.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/example11.png" style="width: 571px; height: 151px;" />
<pre>
<strong>Input:</strong> root = [5,4,9,1,10,null,7]
<strong>Output:</strong> [0,0,0,7,7,null,11]
<strong>Explanation:</strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 5 does not have any cousins so its sum is 0.
- Node with value 4 does not have any cousins so its sum is 0.
- Node with value 9 does not have any cousins so its sum is 0.
- Node with value 1 has a cousin with value 7 so its sum is 7.
- Node with value 10 has a cousin with value 7 so its sum is 7.
- Node with value 7 has cousins with values 1 and 10 so its sum is 11.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/01/11/diagram33.png" style="width: 481px; height: 91px;" />
<pre>
<strong>Input:</strong> root = [3,1,2]
<strong>Output:</strong> [0,0,0]
<strong>Explanation:</strong> The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 3 does not have any cousins so its sum is 0.
- Node with value 1 does not have any cousins so its sum is 0.
- Node with value 2 does not have any cousins so its sum is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>

# 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
import queue
class Solution:
    def helper(self, root):
        q = queue.Queue()
        q.put(root)
        q.put(None)
        sum = 0
        sum_d = {}
        level_d = {}
        level = 0
        while not(q.empty()):
            front = q.get()
            if front == None:
                sum_d[level] = sum
                level += 1
                sum = 0
                if not(q.empty()):
                    q.put(None)
            else:
                sum += front.val
                level_d[front] = level
                if front.left != None:
                    q.put(front.left)
                if front.right != None:
                    q.put(front.right)
        return level_d, sum_d
            
    def helper2(self, root, level_d, sum_d):
        if root == None:
            return None
        if root.left != None and root.right != None:
            temp = root.left.val
            root.left.val = sum_d[level_d[root.left]] - root.left.val - root.right.val
            root.right.val = sum_d[level_d[root.right]] - root.right.val - temp
        elif root.right != None:
            root.right.val = sum_d[level_d[root.right]] - root.right.val
        elif root.left != None:
            root.left.val = sum_d[level_d[root.left]] - root.left.val
        self.helper2(root.left, level_d, sum_d)
        self.helper2(root.right, level_d, sum_d)
        return root


    def replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        level_d, sum_d = self.helper(root)
        
        root = self.helper2(root, level_d, sum_d)
        root.val = 0
        return root

        
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