# 2641. Cousins in Binary Tree II

## Description

Given the root of a binary tree, replace the value of each node in the tree with the sum of all its cousins' values.

Two nodes of a binary tree are cousins if they have the same depth with different parents.

Return the root of the modified tree.

Note that the depth of a node is the number of edges in the path from the root node to it.

Example 1:

Input: root = [5,4,9,1,10,null,7]
Output: [0,0,0,7,7,null,11]
Explanation: 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.


Example 2:

Input: root = [3,1,2]
Output: [0,0,0]
Explanation: 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.


Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 1 <= Node.val <= 104

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private List<Integer> s = new ArrayList<>();

public TreeNode replaceValueInTree(TreeNode root) {
dfs1(root, 0);
root.val = 0;
dfs2(root, 1);
return root;
}

private void dfs1(TreeNode root, int d) {
if (root == null) {
return;
}
if (s.size() <= d) {
}
s.set(d, s.get(d) + root.val);
dfs1(root.left, d + 1);
dfs1(root.right, d + 1);
}

private void dfs2(TreeNode root, int d) {
if (root == null) {
return;
}
int l = root.left == null ? 0 : root.left.val;
int r = root.right == null ? 0 : root.right.val;
if (root.left != null) {
root.left.val = s.get(d) - l - r;
}
if (root.right != null) {
root.right.val = s.get(d) - l - r;
}
dfs2(root.left, d + 1);
dfs2(root.right, d + 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* replaceValueInTree(TreeNode* root) {
vector<int> s;
function<void(TreeNode*, int)> dfs1 = [&](TreeNode* root, int d) {
if (!root) {
return;
}
if (s.size() <= d) {
s.push_back(0);
}
s[d] += root->val;
dfs1(root->left, d + 1);
dfs1(root->right, d + 1);
};
function<void(TreeNode*, int)> dfs2 = [&](TreeNode* root, int d) {
if (!root) {
return;
}
int l = root->left ? root->left->val : 0;
int r = root->right ? root->right->val : 0;
if (root->left) {
root->left->val = s[d] - l - r;
}
if (root->right) {
root->right->val = s[d] - l - r;
}
dfs2(root->left, d + 1);
dfs2(root->right, d + 1);
};
dfs1(root, 0);
root->val = 0;
dfs2(root, 1);
return root;
}
};

• # 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]:
def dfs1(root, d):
if root is None:
return
if len(s) <= d:
s.append(0)
s[d] += root.val
dfs1(root.left, d + 1)
dfs1(root.right, d + 1)

def dfs2(root, d):
if root is None:
return
t = (root.left.val if root.left else 0) + (
root.right.val if root.right else 0
)
if root.left:
root.left.val = s[d] - t
if root.right:
root.right.val = s[d] - t
dfs2(root.left, d + 1)
dfs2(root.right, d + 1)

s = []
dfs1(root, 0)
root.val = 0
dfs2(root, 1)
return root


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func replaceValueInTree(root *TreeNode) *TreeNode {
s := []int{}
var dfs1 func(*TreeNode, int)
dfs1 = func(root *TreeNode, d int) {
if root == nil {
return
}
if len(s) <= d {
s = append(s, 0)
}
s[d] += root.Val
dfs1(root.Left, d+1)
dfs1(root.Right, d+1)
}
var dfs2 func(*TreeNode, int)
dfs2 = func(root *TreeNode, d int) {
if root == nil {
return
}
l, r := 0, 0
if root.Left != nil {
l = root.Left.Val
}
if root.Right != nil {
r = root.Right.Val
}
if root.Left != nil {
root.Left.Val = s[d] - l - r
}
if root.Right != nil {
root.Right.Val = s[d] - l - r
}
dfs2(root.Left, d+1)
dfs2(root.Right, d+1)
}
dfs1(root, 0)
root.Val = 0
dfs2(root, 1)
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function replaceValueInTree(root: TreeNode | null): TreeNode | null {
const s: number[] = [];
const dfs1 = (root: TreeNode | null, d: number): void => {
if (!root) {
return;
}
if (s.length <= d) {
s.push(0);
}
s[d] += root.val;
dfs1(root.left, d + 1);
dfs1(root.right, d + 1);
};
const dfs2 = (root: TreeNode | null, d: number): void => {
if (!root) {
return;
}
const t = (root.left?.val ?? 0) + (root.right?.val ?? 0);
if (root.left) {
root.left.val = s[d] - t;
}
if (root.right) {
root.right.val = s[d] - t;
}
dfs2(root.left, d + 1);
dfs2(root.right, d + 1);
};
dfs1(root, 0);
root.val = 0;
dfs2(root, 1);
return root;
}