<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
class Solution:
    def build_level_maps(self, node, parent, level, level_map, parent_map, level_node_map, orig_vals):
        if node is None:
            return
        parent_map[node] = parent
        level_node_map[node] = level
        level_map[level] = level_map.get(level, 0)+node.val
        orig_vals[node] = node.val
        self.build_level_maps(node.left, node, level+1, level_map, parent_map, level_node_map, orig_vals)
        self.build_level_maps(node.right, node, level+1, level_map, parent_map,level_node_map, orig_vals)


    def get_siblings_sum(self, parent, node,         orig_vals ):
        if parent is None:
            return orig_vals[node]
        return orig_vals[parent.left]  + orig_vals[parent.right]
    def replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        curr_work = [root]
        orig_vals = {}
        orig_vals[None] = 0
        level_sum_map = {}
        level_node_map = {}
        parent_map = {}

        self.build_level_maps(root, None, 0, level_sum_map, parent_map, level_node_map, orig_vals)
        while curr_work:
            curr_node = curr_work.pop()
            if curr_node.left is not None:
                curr_work.append(curr_node.left)
            if curr_node.right is not None:
                curr_work.append(curr_node.right)
            curr_node.val = level_sum_map[level_node_map[curr_node]] - self.get_siblings_sum(parent_map[curr_node],curr_node, orig_vals)
        return root

        
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