<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 replaceValueInTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        q = collections.deque()
        q.append([root, None])

        while q:
            l = len(q)
            total = 0
            tmp = [] 
            for _ in range(l):
                cur = q.popleft()
                total += cur[0].val
                if cur[0].left:
                    q.append([cur[0].left, cur[0]])
                if cur[0].right:
                    q.append([cur[0].right, cur[0]])
                tmp.append(cur)
            for i in range(len(tmp)):
                t = total - tmp[i][0].val
                if i + 1 < len(tmp) and tmp[i+1][1] == tmp[i][1]:
                    #print(tmp[i][0].val, t, tmp[i+1][0].val)
                    tmp[i][0].val = t - tmp[i+1][0].val
                elif i - 1 >=0 and tmp[i-1][1] == tmp[i][1]:
                    #print(tmp[i][0].val, t, tmp[i-1][0].val)
                    tmp[i][0].val = tmp[i-1][0].val
                else:
                    #print(tmp[i][0].val, t)
                    tmp[i][0].val = t
        return root
                    
                
            

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