LeetCode：Populating Next Right Pointers in Each Node I,II 编程之美3.10

解答：

        队列实现按排遍历

class Solution {public:struct InvoLevelNode {    TreeLinkNode *node;    int level;    InvoLevelNode(TreeLinkNode *p,int n):node(p),level(n){}    InvoLevelNode(){}//重定义构造函数时，一定要把程序里面出现情况全部构造完全};    void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(root ==NULL)return ;        deque<InvoLevelNode>L;        int level = 1;        InvoLevelNode p,q;        L.push_back(InvoLevelNode(root,1));        while(L.size())        {            p = L.front();            L.pop_front();            if(p.node->left)               L.push_back(InvoLevelNode(p.node->left,p.level+1));            if(p.node->right)               L.push_back(InvoLevelNode(p.node->right,p.level+1));            if(L.size())                q= L.front();            else            {                p.node->next = NULL;                continue;            }            if(p.level == q.level)               p.node->next = q.node;            else               p.node->next = NULL;        }                            }};

查看编程之美 3.10 分层遍历二叉树感觉代码更好点 如下

class Solution {public:    void connect(TreeLinkNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(root==NULL)return;        vector<TreeLinkNode*> V;        TreeLinkNode* node;        int last = 1;        int nowp = 0;        V.push_back(root);        while(last!= nowp)        {            node = V[nowp];            if(node->left)             //这个要在last更新之前做               V.push_back(node->left);            if(node->right)               V.push_back(node->right);            if(nowp == last - 1 )            {               node->next = NULL;                  last = V.size();//遍历完一层，新压入的只有可能是下一层的元素            }            else            {                node->next = V[nowp+1];            }                       nowp++;        }    }};