2415. Reverse Odd Levels of Binary Tree

Description

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

• For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].

Return the root of the reversed tree.

A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.

The level of a node is the number of edges along the path between it and the root node.

 

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation: 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation: 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation: 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

 

Constraints:

• The number of nodes in the tree is in the range [1, 214].
• 0 <= Node.val <= 105
• root is a perfect binary tree.

 

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 reverseOddLevels = function (root) {
  let queue = [root];
  let level = 0;

  const reverseNodesVal = nodes => {
    if (level % 2 === 0) return;
    const n = nodes.length;

    for (let index = 0; index < n / 2; index++) {
      const nodeA = nodes[index];
      const nodeB = nodes[n - index - 1];

      [nodeA.val, nodeB.val] = [nodeB.val, nodeA.val];
    }
  };

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

    level += 1;

    for (const node of queue) {
      if (node.left) {
        nextQueue.push(node.left);
      }
      if (node.right) {
        nextQueue.push(node.right);
      }
    }
    reverseNodesVal(nextQueue);
    queue = nextQueue;
  }
  return root;
};