117. Populating Next Right Pointers in Each Node II

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
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low一点的方法：同116 Populating Next Right Pointers in Each Node，的方法一一样，采用层次遍历的思想来解决，但运行时间长点，代码:

void connect(TreeLinkNode *root) {         if(NULL == root)            return;        queue<TreeLinkNode*> que;        que.push(root);        int curLenNode = 1;        TreeLinkNode* pre = NULL;        while(!que.empty())        {            TreeLinkNode *temp = que.front();            que.pop();            curLenNode --;            if(pre != NULL)                pre->next = temp;            pre = temp;            if(temp->left)                que.push(temp->left);            if(temp->right)                que.push(temp->right);            if(curLenNode == 0)            {                pre = NULL; //注意在每一层遍历结束，都要将pre设置为NULL，否则Pre保留的是上一层的最后一个节点，故而会有问题                temp->next = NULL;//最每一层的最后会将当前层的最后一个节点的next赋值为NULL                curLenNode = que.size();            }        }    }

更好的方法值得推敲：

public void connect(TreeLinkNode root) {        TreeLinkNode head = null; //head of the next level        TreeLinkNode prev = null; //the leading node on the next level        TreeLinkNode cur = root;  //current node of current level        while (cur != null) {            while (cur != null) { //iterate on the current level                //left child                if (cur.left != null) {                    if (prev != null) {                        prev.next = cur.left;                    } else {                        head = cur.left;                    }                    prev = cur.left;                }                //right child                if (cur.right != null) {                    if (prev != null) {                        prev.next = cur.right;                    } else {                        head = cur.right;                    }                    prev = cur.right;                }                //move to next node                cur = cur.next;            }            //move to next level            cur = head;            head = null;            prev = null;        }    }

更好的方法二：
https://discuss.leetcode.com/topic/18119/simple-40ms-c-o-n-o-1-solution-with-only-one-while-loop