<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
from collections import Counter
class Solution:
    def replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        levels = Counter()
        def dfs(root, depth):
            if not root:
                return
            levels[depth] += root.val
            dfs(root.left, depth + 1)
            dfs(root.right, depth + 1)
        
        dfs(root, 0)
        # print(levels)
        def dfs1(root, depth):
            if not root:
                return
            root.val = levels[depth] - root.val
            children = (root.left.val if root.left else 0) + (root.right.val if root.right else 0)
            if root.left:
                root.left.val = children
            if root.right:
                root.right.val = children
            dfs1(root.left, depth + 1)
            dfs1(root.right, depth + 1)
        
        
        def dfs2(root, depth, sibling):
            if not root:
                return
            children = (root.left.val if root.left else 0) + (root.right.val if root.right else 0)
            root.val = levels[depth] - sibling
            dfs2(root.left, depth + 1, children)
            dfs2(root.right, depth + 1, children)

        dfs2(root, 0, root.val) 
        # root.val = 0
        return root

        
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