116. Populating Next Right Pointers in Each Node Add to List

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
解题思路：一开始没看到要用O(1)的空间复杂度，用了个vector将所有遍历经过的点都存起来了。这是适用于任意树形，但是空间复杂度为O(n).

/**
 * 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 dfs(TreeLinkNode *root,int level){
        if(!root)return;
        if(level==res.size()){
            res.push_back(vector<TreeLinkNode*>());
        }
        res[level].push_back(root);
        dfs(root->left,level+1);
        dfs(root->right,level+1);
    }
    void connect(TreeLinkNode *root) {
        dfs(root,0);
       // cout<<res.size()<<endl;
        for(int i=0;i<res.size();i++){
            for(int j=0;j<res[i].size();j++)
                if(j==res[i].size()-1) res[i][j]->next=NULL;
                else res[i][j]->next=res[i][j+1];
        }
    }
private:
    vector<vector<TreeLinkNode* > >res;
};

看清题目之后，以为树是CBT,所以就有了以下更简便的方法，利用next可以容易的得到右节点的指向

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

 还有一种是用两个指针来模拟队列