LeetCode Populating Next Right Pointers in Each Node & Populating Next Right Pointers in Each Node I

Populating Next Right Pointers in Each Node

Description:

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.

Solution:

Just a bfs here is OK.

It is like go through the tree level by level, so we can this is kind of like a level-order. And for each level, I try to relink the nodes here.

import java.util.Iterator;import java.util.LinkedList;public class Solution {public void connect(TreeLinkNode root) {if (root == null)return;LinkedList<TreeLinkNode> currentList = new LinkedList<TreeLinkNode>();LinkedList<TreeLinkNode> followingList;currentList.add(root);while (true) {followingList = new LinkedList<TreeLinkNode>();while (!currentList.isEmpty()) {TreeLinkNode node = currentList.poll();if (node.left != null)followingList.add(node.left);if (node.right != null)followingList.add(node.right);}if (followingList.isEmpty())break;TreeLinkNode pre = followingList.getFirst();for (Iterator<TreeLinkNode> ite = followingList.iterator(); ite.hasNext();) {TreeLinkNode cur = ite.next();pre.next = cur;pre = cur;}currentList = followingList;}}}

Populating Next Right Pointers in Each Node II

Description:

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Solution 1:

The #1 solution is like the solution to the former problem. But get a MLE.

import java.util.Iterator;import java.util.LinkedList;/** * Definition for binary tree with next pointer. public class TreeLinkNode { int * val; TreeLinkNode left, right, next; TreeLinkNode(int x) { val = x; } } */public class Solution {public void connect(TreeLinkNode root) {if (root == null)return;LinkedList<TreeLinkNode> currentList = new LinkedList<TreeLinkNode>();LinkedList<TreeLinkNode> followingList;currentList.add(root);while (true) {followingList = new LinkedList<TreeLinkNode>();while (!currentList.isEmpty()) {TreeLinkNode node = currentList.poll();if (node.left != null)followingList.add(node.left);if (node.right != null)followingList.add(node.right);}if (followingList.isEmpty())break;TreeLinkNode pre = followingList.getFirst();for (Iterator<TreeLinkNode> ite = followingList.iterator(); ite.hasNext();) {TreeLinkNode cur = ite.next();pre.next = cur;pre = cur;}currentList = followingList;}}}

Solution 2:

To avoid the MLE here, we can search the tree in this way. For each level, we search from right to left, so that we do not need to use a linkedlist to record all the nodes here.

import java.util.Iterator;import java.util.LinkedList;/** * Definition for binary tree with next pointer. public class TreeLinkNode { int * val; TreeLinkNode left, right, next; TreeLinkNode(int x) { val = x; } } */public class Solution {public void connect(TreeLinkNode root) {if (root == null)return;LinkedList<TreeLinkNode> list = new LinkedList<TreeLinkNode>();LinkedList<Integer> levels = new LinkedList<Integer>();list.add(root);levels.add(0);int current_level = 0;TreeLinkNode temp;TreeLinkNode pre = null;while (!list.isEmpty()) {temp = list.poll();int l = levels.poll();if (current_level != l) {current_level = l;pre = null;}temp.next = pre;pre = temp;if (temp.right != null) {list.add(temp.right);levels.add(current_level + 1);}if (temp.left != null) {list.add(temp.left);levels.add(current_level + 1);}}}}

Final Solution for both:

For the level order traversal, we can have one more promoted version. And if we can use this version, the solution to the two problems are much simpler.

import java.util.LinkedList;public class Solution {public void connect(TreeLinkNode root) {if (root == null)return;LinkedList<TreeLinkNode> list = new LinkedList<TreeLinkNode>();list.add(root);TreeLinkNode node, next;int len;while (!list.isEmpty()) {next = null;len = list.size();for (int i = 0; i < len; i++) {node = list.poll();node.next = next;next = node;if (node.right != null)list.add(node.right);if (node.left != null)list.add(node.left);}}}}class TreeLinkNode {int val;TreeLinkNode left, right, next;public TreeLinkNode() {// TODO Auto-generated constructor stub}}