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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:
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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//二叉树进行层次遍历
//判断最底层的条件：如果当前访问的层有任一节点存在子节点，说明当前层不是最底层。
class Solution{
public:
TreeLinkNode* nextLeft;
bool hasToTheEnd;
void connect(TreeLinkNode *root)
{
if (!root) return;
hasToTheEnd = false;
int level = 1;
while (!hasToTheEnd)
{
hasToTheEnd= true;
nextLeft= NULL;
VisitLevel(root, level);
++level;
}
}
void VisitLevel(TreeLinkNode* node, int level)
{
if (level == 1)
{
if (nextLeft != NULL)
{
nextLeft->next = node;
}
nextLeft = node;
if (node->left != NULL || node->right != NULL)
{
hasToTheEnd= false;
}
}
if (node->left) VisitLevel(node->left, level - 1);
if (node->right) VisitLevel(node->right, level - 1);
}
};
LeetCode 117:Populating Next Right Pointers in Each Node II
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原文地址：http://blog.csdn.net/geekmanong/article/details/50439036
