Populating Next Right Pointers in Each Node and II[LeetCode]

Populating Next Right Pointers in Each Node

 

Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1       /  \      2    3     / \  / \    4  5  6  7

After calling your function, the tree should look like:

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

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

两题都采用层次遍历的方法，都可实现，一份代码

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  TreeLinkNode *left, *right, *next; *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} * }; */class Solution {public:    void connect(TreeLinkNode *root) {        if(root == NULL) return;        queue<TreeLinkNode *> qu;        qu.push(root);        root->next = NULL;        TreeLinkNode * pre;        TreeLinkNode * cur;        int num =1,n=0;        while(!qu.empty()){            pre = qu.front();            qu.pop();            n = 0;            if(pre->left){qu.push(pre->left);n++;}            if(pre->right){qu.push(pre->right);n++;}                       while(num--){                if(num == 0){                    pre->next = NULL;                }                else{                    cur = qu.front();                    qu.pop();                    if(cur->left){qu.push(cur->left);n++;}                    if(cur->right){qu.push(cur->right);n++;}                    pre->next = cur;                    pre = cur;                }            }            num = n;                    }    }};