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

Analysis: This problem is level-order traversal deformation, and level-order traversal different is that the problem requires constant space. General Level-order traversal with a container to save the previous layer of the node is O (n) space complexity, so here are different direct copy. To analyze this problem, a node P's left child's next is node P's right child, p's node right child's next is P's next node left child, by this we can only maintain node p to complete populate. The time complexity is O (n) and the spatial complexity is O (1). The code is as follows:

classSolution { Public:    voidConnect (Treelinknode *root) {        if(Root = NULL)return; Treelinknode*head = root;//Point-to- head of every level         while(Head->left && head->Right ) {Treelinknode*cur = head;//iterate through each node at a level             while(cur) {cur->left->next = cur->Right ; Cur->right->next = cur->next?cur->next->Left:null; Cur= cur->Next; } head= head->left;//update head to left child of previous head        }    }};

Leetcode:populating Next Right pointers in each Node