<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>

class Solution:
  def replaceValueInTree(self, root: TreeNode | None) -> TreeNode | None:
    levelSums = []

    def dfs(root: TreeNode | None, level: int) -> None:
      if not root:
        return
      if len(levelSums) == level:
        levelSums.append(0)
      levelSums[level] += root.val
      dfs(root.left, level + 1)
      dfs(root.right, level + 1)

    def replace(
        root: TreeNode | None,
        level: int, curr: TreeNode | None,
    ) -> TreeNode | None:
      nextLevel = level + 1
      nextLevelCousinsSum = (
          (levelSums[nextLevel] if nextLevel < len(levelSums) else 0) -
          (root.left.val if root.left else 0) -
          (root.right.val if root.right else 0))
      if root.left:
        curr.left = TreeNode(nextLevelCousinsSum)
        replace(root.left, level + 1, curr.left)
      if root.right:
        curr.right = TreeNode(nextLevelCousinsSum)
        replace(root.right, level + 1, curr.right)
      return curr

    dfs(root, 0)
    return replace(root, 0, TreeNode(0))
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