[leetcode][tree][dfs] Populating Next Right Pointers in Each Node

题目：

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

/** * 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://注意题目中说明输入是满二叉树，用先序遍历，关键是利用上一层已经求出的next域//root的左孩子的next是右孩子，root的右孩子的next是root->next的左孩子    void connect(TreeLinkNode *root) {        if(NULL == root || NULL == root->left && NULL == root->right) return;//叶节点没有左右孩子，直接返回        root->left->next = root->right;        root->right->next = root->next ? root->next->left : NULL;        connect(root->left);        connect(root->right);    }};

按层遍历实现：

/** * 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://按层遍历TreeLinkNode *curLevelFirst = NULL;    void connect(TreeLinkNode *root) {        if(NULL == root || NULL == root->left) return;        if(NULL == curLevelFirst) curLevelFirst = root;        root->left->next = root->right;        root->right->next = root->next ? root->next->left : NULL;        if(root->next) connect(root->next);        else{            curLevelFirst = curLevelFirst->left;            connect(curLevelFirst);        }    }};