56116. Populating Next Right Pointers in Each Node

Posted 2020-08-30 会咬人的兔子

116. Populating Next Right Pointers in Each Node

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• Total Accepted: 118451
• Total Submissions: 321021
• Difficulty: Medium
• Contributors: Admin

 

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

Solution 1：添加空指针来分层

/**
 * 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) {
        queue<TreeLinkNode*> q;
        if(!root) return;
        q.push(root);
        q.push(NULL);//比层序遍历多了要push NULL进去
        //while(!q.empty()){
        while(true){
            //int size = q.size();
            //for(int i = 0; i < size; i++){
                TreeLinkNode* tmp = q.front();
                q.pop();
                if(tmp){
                    tmp -> next = q.front();
                    if(tmp -> left) q.push(tmp -> left);
                    if(tmp -> right) q.push(tmp -> right);
                }else{
                    if(q.size() == 0 || q.front() == NULL) return;//添加空指针NULL来达到分层的目的，使每层的最后一个节点的next可以指向NULL
                    q.push(NULL);
                }
            //}
        }
    }
};

Solution 2：记录size来分层

/**
 * 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) {
        queue<TreeLinkNode*> q;
        if(!root) return;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; i++){
                TreeLinkNode* tmp = q.front();
                q.pop();
                
                if(i < size - 1){//给直到每层的倒数第二个node设置next指针
                    tmp -> next = q.front();
                }
                //tmp -> next = q.front();
                
                if(tmp -> left) q.push(tmp -> left);
                if(tmp -> right) q.push(tmp -> right);
            }
        }
    }
};

Solution 3：Recursion

/**
 * 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 -> right){
            root -> right -> next = root -> next ? root -> next -> left : NULL;//需要判断root -> next是否为空，否则空指针是没有左指针的
        }
        connect(root -> left);
        connect(root -> right);
    }
};