2641. Cousins in Binary Tree II

Problem Statement

Medium

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

Java — 1pass BFS

Source file

/**
 * 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 {
    public TreeNode replaceValueInTree(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int levelCount = 1, levelSum = root.val;
        while (!q.isEmpty()) {
            int nextLevelSum = 0;
            for (int i = 0; i < levelCount; i++) {
                TreeNode front = q.poll();
                front.val = levelSum - front.val;
                int siblingSum = (front.left != null ? front.left.val : 0)
                        + (front.right != null ? front.right.val : 0);
                nextLevelSum += siblingSum;
                if (front.left != null) {
                    front.left.val = siblingSum;
                    q.offer(front.left);
                }
                if (front.right != null) {
                    front.right.val = siblingSum;
                    q.offer(front.right);
                }
            }
            levelSum = nextLevelSum;
            levelCount = q.size();
        }
        return root;
    }
}

Java — 2pass BFS+DFS

Source file

/**
 * 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 {
    public TreeNode replaceValueInTree(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        ArrayList<Integer> levelSums = new ArrayList<>();
        q.offer(root);
        int levelCount = 1;
        while (!q.isEmpty()) {
            int levelSum = 0;
            for (int i = 0; i < levelCount; i++) {
                TreeNode front = q.poll();
                levelSum += front.val;
                if (front.left != null)
                    q.offer(front.left);
                if (front.right != null)
                    q.offer(front.right);
            }
            levelSums.add(levelSum);
            levelCount = q.size();
        }
        // System.out.println(levelSums);
        modify(root, 0, root.val, levelSums);
        return root;
    }

    private void modify(TreeNode root, int depth, int siblingsSum, List<Integer> levelSums) {
        if (root == null)
            return;
        root.val = levelSums.get(depth) - siblingsSum;
        int childrenSum = (root.left != null ? root.left.val : 0) + (root.right != null ? root.right.val : 0);
        modify(root.left, depth + 1, childrenSum, levelSums);
        modify(root.right, depth + 1, childrenSum, levelSums);
    }
}

Java

Source file

/**
 * 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 {
    public TreeNode replaceValueInTree(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int levelCount = 1, levelSum = root.val;
        while (!q.isEmpty()) {
            int nextLevelSum = 0;
            for (int i = 0; i < levelCount; i++) {
                TreeNode front = q.poll();
                front.val = levelSum - front.val;
                int siblingSum = (front.left != null ? front.left.val : 0)
                        + (front.right != null ? front.right.val : 0);
                nextLevelSum += siblingSum;
                if (front.left != null) {
                    front.left.val = siblingSum;
                    q.offer(front.left);
                }
                if (front.right != null) {
                    front.right.val = siblingSum;
                    q.offer(front.right);
                }
            }
            levelSum = nextLevelSum;
            levelCount = q.size();
        }
        return root;
    }
}