leetcode-116-Populating Next Right Pointers in Each Node

                             Populating Next Right Pointers in Each Nod

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

递归求解，

root->left->next指向root->right,

root->right->next指向root->next->left，前提是root->next存在，所以递归时，要先求root->right，再root->left

/** * 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) {   //C++        if(!root) return ;        if(!root->left) return;                root->left->next=root->right;        if(root->next) root->right->next=root->next->left;                connect(root->right); // 先root->right        connect(root->left);    }};

优化

• You may only use constant extra space.

使用常数级的额外空间，故不能用深搜和广搜。

但每一层的节点都通过next指针连起来，故可以以每层第一个节点开始，往后依次遍历

/** * Definition for binary tree with next pointer. * struct TreeLinkNode { *  int val; *  struct TreeLinkNode *left, *right, *next; * }; * */void connect(struct TreeLinkNode *root) { // C        if(!root) return ;        if(!root->left) return;                struct TreeLinkNode* node;        while(root->left){  // root 为 每一层的第一个节点            node=root;            while(node){   // node 遍历每一层节点  并找到其左右孩子的next                node->left->next=node->right;                if(node->next)                      node->right->next=node->next->left;                node=node->next;            }            root=root->left;        }}