Populating Next Right Pointers in Each Node ---LeetCode

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

解题思路：

观察可以发现，每一层从左至右每一个节点依次指向它右边的节点。因此可以用层序遍历 Binary Tree Level Order Traversal 来解决这道题，在遍历到这一层的时候，判断它是不是本层最右的元素，如果不是就指向旁边的元素，如果是就指向 null。

在层序遍历中，我们维护一个队列来依次将本层元素入列。那么要如何判断本层元素所处的位置呢？因此可以再维护一个队列，里面装载着层数信息。

所以这道题用两个队列解决，应该是最直接的解法。一个装载节点，一个装载层数信息。

/** * 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 ;        Queue<TreeLinkNode> nodeQ = new LinkedList<>();        Queue<Integer> depthQ = new LinkedList<>();        if (root != null) {            nodeQ.offer(root);            depthQ.offer(1);        }        while (!nodeQ.isEmpty()) {            TreeLinkNode node = nodeQ.poll();            int depth = depthQ.poll();            if (depthQ.isEmpty()) {                node.next = null;            } else if (depthQ.peek() > depth) {                node.next = null;            } else {                node.next = nodeQ.peek();            }            if (node.left != null) {                nodeQ.offer(node.left);                depthQ.offer(depth + 1);            }            if (node.right != null) {                nodeQ.offer(node.right);                depthQ.offer(depth + 1);            }        }    }}

另外一种在 Program Creek 里学到的解法：可以不用队列。仅仅维护四个指针，其中两个指针指向两层的第一个节点，另外两个节点用于遍历层。

/** * 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 ;        TreeLinkNode lastHead = root; // previous level's head        TreeLinkNode lastCurr = null; // previous level's pointer        TreeLinkNode currHead = null; // current level's head        TreeLinkNode current  = null; // current level's pointer        while (lastHead != null) {            lastCurr = lastHead;            while (lastCurr != null) {                if (currHead == null) {                    currHead = lastCurr.left;                    current  = lastCurr.left;                } else {                    current.next = lastCurr.left;                    current = current.next;                }                if (currHead != null) {                    current.next = lastCurr.right;                    current = current.next;                }                lastCurr = lastCurr.next;            }            lastHead = currHead;            currHead = null;        }    }}