116. Populating Next Right Pointers in Each Node

Problem:

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

     1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

Solution:

ONE

Recursive.

For every node:

• Left child: points to node.right.
• Right child: points to node.next.left if node.next exists.

/**
 * Definition for binary tree with next pointer.
 * function TreeLinkNode(val) {
 *     this.val = val;
 *     this.left = this.right = this.next = null;
 * }
 */

/**
 * @param {TreeLinkNode} root
 * @return {void} Do not return anything, modify tree in-place instead.
 */
let connect = function (root) {
  if (!root) {
    return;
  }
  if (root.left !== null) {
    root.left.next = root.right;
    connect(root.left);
  }
  if (root.right !== null) {
    if (root.next !== null) {
      root.right.next = root.next.left;
    }
    connect(root.right);
  }
};

TWO

Level order traversal.

/**
 * Definition for binary tree with next pointer.
 * function TreeLinkNode(val) {
 *     this.val = val;
 *     this.left = this.right = this.next = null;
 * }
 */

/**
 * @param {TreeLinkNode} root
 * @return {void} Do not return anything, modify tree in-place instead.
 */
let connect = function (root) {
  if (!root) {
    return;
  }

  const queue = [NaN, root];
  while (queue.length > 1) {
    const node = queue.shift();
    if (node !== node) {
      for (let i = 0; i < queue.length; i++) {
        queue[i].next = queue[i + 1] || null;
      }
      queue.push(NaN);
    } else {
      if (node.left !== null) {
        queue.push(node.left);
      }
      if (node.right !== null) {
        queue.push(node.right);
      }
    }
  }
};

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