leetcode 116填充每个节点的下一个右侧节点指针

time O(n) ，sapce O(n)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() {}

    Node(int _val, Node* _left, Node* _right, Node* _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
/**
层序遍历，然后每一层左边节点指向右边节点，最后一个节点指向NULL
**/
class Solution {
public:
    Node* connect(Node* root) {
        if(root==NULL||root->left==NULL) return root;
        queue<Node*>q;
        q.push(root->left),q.push(root->right);
        while(!q.empty()){
            int k=q.size();
            Node* l,*r=q.front();
            q.pop();
            if(r->left) q.push(r->left),q.push(r->right);
            for(int i=1;i<k;i++){
                l=r,r=q.front();q.pop();
                l->next=r;
                if(r->left) q.push(r->left),q.push(r->right);
            }
        }
        return root;
    }
};

原文地址：https://www.cnblogs.com/joelwang/p/10920448.html