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

Solution: Breadth-First Search

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
const replaceValueInTree = function (root) {
  let queue = [{ node: root, parent: new TreeNode() }];
  let depthSum = 0;

  while (queue.length) {
    const nextQueue = [];
    let nextDepthSum = 0;

    for (const { node, parent } of queue) {
      const childSum = (node.left?.val ?? 0) + (node.right?.val ?? 0);

      node.val = depthSum - (parent.childSum ?? 0);
      node.childSum = childSum;
      nextDepthSum += childSum;
      if (node.left) nextQueue.push({ node: node.left, parent: node });
      if (node.right) nextQueue.push({ node: node.right, parent: node });
    }
    depthSum = nextDepthSum;
    queue = nextQueue;
  }
  return root;
};