[LeetCode]Populating Next Right Pointers in Each Node II

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

利用宽搜顺序搜索每一行的二叉树，然后利用深搜搜索该节点的下一个next节点。

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {Queue<TreeLinkNode> que = new LinkedList<>();public void connect(TreeLinkNode root) {if (root == null)return;que.offer(root);while (!que.isEmpty()) {TreeLinkNode tln = que.poll();if (tln.left != null) {que.offer(tln.left);tln.left.next = dfs2(tln);}if (tln.right != null) {que.offer(tln.right);tln.right.next = dfs3(tln.next);}}}private TreeLinkNode dfs2(TreeLinkNode root){if(root == null) return null;if(root.right!=null) return root.right;if(root.next!=null&&root.next.left!=null){return root.next.left;}return dfs2(root.next);}private TreeLinkNode dfs3(TreeLinkNode root){if(root ==null) {return null;}else{if(root.left!=null) return root.left;if(root.right!=null)return root.right;}return dfs3(root.next);}}

递归：参考 戴方勤 LeetCode题解

/** * 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 dummy = new TreeLinkNode(-1);for(TreeLinkNode cur=root,fut=dummy;cur!=null;cur=cur.next){if(cur.left!=null){fut.next = cur.left;fut = fut.next;}if(cur.right!=null){fut.next = cur.right;fut = fut.next;}}connect(dummy.next);}}