[Leetcode] Populating Next right pointers in each Node

Given a binary tree

    struct Treelinknode {      treelinknode *left;      Treelinknode *right;      Treelinknode *next;    }

Populate each of the next pointer to the next right node. If There is no next right node, the next pointer should are set to NULL .

Initially, all next pointers is set to NULL .

Note:

• Constant extra space.
• You could assume that it was a perfect binary tree (ie, all leaves was at the same level, and every parent had both children).

For example,
Given the following perfect binary tree,

         1       /        2    3     /\  /     4  5  6  7

After calling your function, the tree is should look like:

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

Wide Search + recursion

Recursion two times

The tree after CONNECTLR ()

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

Connectrl () for connecting right subtree to left subtree run time: 436ms

/** * 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;CONNECTLR (root); Connectrl (ro OT);} private void Connectlr (Treelinknode root) {if (root.left==null) Return;root.left.next = ROOT.RIGHT;CONNECTLR (root.left ); CONNECTLR (root.right);} private void Connectrl (Treelinknode root) {if (root.left==null) return;if (root.next!=null) {Root.right.next = Root.next.left;} Connectrl (Root.left); Connectrl (root.right);}}

Time to complete a recursive run: 440ms

/** * 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) {connect (root,null);} private void Connect (Treelinknode root, treelinknode brother) {if (root = null) Return;root.next = Brother;connect ( Root.left,root.right); if (brother!=null) {connect (root.right,brother.left);} Else{connect (Root.right,null);}}}

[Leetcode] Populating Next right pointers in each Node