Populating Next Right Pointers in Each Node II

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Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \    \    4-> 5 -> 7 -> NULL

点击打开原题链接

基本上是层次遍历稍微修改下就可以了，算法比较简单，注意下边界条件即可，也可以用把每一层放入vector<vector<> >中去，但是会开辟新的内存，这里

直接判断是否在一层上，代码稍微复杂一点点。

struct TreeLinkNode  {  int val;  TreeLinkNode *left, *right, *next;  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}}; struct node { int level; TreeLinkNode* TLN; };class Solution {public:void connect(TreeLinkNode *root) {if (root == NULL){return;}deque<node> deque_TreeLinkNode;node pre,current;pre.level = 0;pre.TLN = root;deque_TreeLinkNode.push_back(pre);pre.TLN = NULL;while(!deque_TreeLinkNode.empty()){node temp = deque_TreeLinkNode.front();if (temp.TLN->left != NULL){node left;left.level = temp.level+1;left.TLN = temp.TLN->left;deque_TreeLinkNode.push_back(left);}if (temp.TLN->right != NULL){node right;right.level = temp.level+1;right.TLN = temp.TLN->right;deque_TreeLinkNode.push_back(right);}current= deque_TreeLinkNode.front();deque_TreeLinkNode.pop_front();if (pre.TLN == NULL){pre.TLN = current.TLN;pre.level = current.level;//deque_TreeLinkNode.pop_front();continue;}else{if (current.level == pre.level){pre.TLN->next = current.TLN;pre=current;continue;//current->TLN = NULL;//current->level = 0;}else{pre.TLN->next = NULL;pre = current;}}}pre.TLN->next = NULL;}private:};

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