LeetCode 117 Populating Next Right Pointers in Each Node II

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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?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

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

Anwser 1：

public void connect(TreeLinkNode root) {if (null == root) {return;}TreeLinkNode cur = root.next;TreeLinkNode p = null;while (cur != null) { // find last right node (left or right)if (cur.left != null) {p = cur.left;break;}if (cur.right != null) {p = cur.right;break;}cur = cur.next;}if (root.right != null) {root.right.next = p;}if (root.left != null) {root.left.next = root.right != null ? root.right : p;}connect(root.right); // from right to leftconnect(root.left);}

注意点：
1） list为非完美二叉树，右分支可能为空，因此从right -> left 遍历
2）从最右分支开始查找，且root没有 left 节点，则找 right 节点
Anwser 2：

public void connect(TreeLinkNode root) {if (null == root) {return;}LinkedList<TreeLinkNode> Q = new LinkedList<TreeLinkNode>(); // save one// line// root(s)LinkedList<TreeLinkNode> Q2 = new LinkedList<TreeLinkNode>();; // save next one line root(s), swap with QQ.push(root);while (!Q.isEmpty()) {TreeLinkNode tmp = Q.getFirst();Q.pop();if (tmp.left != null)Q2.add(tmp.left);if (tmp.right != null)Q2.add(tmp.right);if (Q.isEmpty()) {tmp.next = null;LinkedList<TreeLinkNode> tmpQ = Q; // swap queueQ = Q2;Q2 = tmpQ;} else {tmp.next = Q.getFirst();}}}

注意点：
1）新增一个Q2队列，保存下一行的全部元素，辅助判断是最后一个元素（Q为空）则置为NULL
2） queue队列实现比递归要好
Anwser 3：

public void connect(TreeLinkNode root) {LinkedList<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();if (root == null)return;TreeLinkNode p;queue.add(root);queue.add(null);// flagwhile (!queue.isEmpty()) {p = queue.pop();if (p != null) {if (p.left != null) {queue.add(p.left);}if (p.right != null) {queue.add(p.right);}p.next = queue.getFirst();} else {if (queue.isEmpty()) {return;}queue.add(null);}}}

Anwser 4:使用计数器

/** * 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) {LinkedList<TreeLinkNode> linkedList = new LinkedList<TreeLinkNode>();if (root == null)return;TreeLinkNode p;int num = 1;linkedList.addLast(root);while (!linkedList.isEmpty()) {int i = 0;int temp = 0;while (i < num) {p = linkedList.pop();if (p.left != null) {linkedList.addLast(p.left);temp++;}if(p.right!=null){linkedList.addLast(p.right);temp++;}if(i!=num-1)p.next=linkedList.getFirst();else p.next=null;i++;}num = temp;}}}

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