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
如何处理各个细节的问题很重要,决定了能否写出更加简洁的程序来。
void connect(TreeLinkNode *root)
{
while (root && !root->left && !root->right) root = root->next;
if (root == NULL) return;
TreeLinkNode *rightSibling;
TreeLinkNode *p1 = root;
while (p1)
{
TreeLinkNode *p_next = p1->next;
while (p_next && !p_next->left && !p_next->right)
p_next = p_next->next;
if (p_next)
rightSibling = p_next->left? p_next->left:p_next->right;
else rightSibling = NULL;
if (p1->left && p1->right)
{
p1->left->next = p1->right;
p1->right->next = rightSibling;
}
else if (p1->right)
{
p1->right->next = rightSibling;
}
else
{
p1->left->next = rightSibling;
}
p1 = p_next;
}
if (root->left)
connect(root->left);
else if(root->right)
connect(root->right);
}Leetcode论坛上的一个简洁的程序,写得太好了,编程功底就在微小的差别中体现出来了。
具体思想是和我上面的程序一样的。
//LeetCode论坛上的
// the link of level(i) is the queue of level(i+1)
void connect2(TreeLinkNode * n) {
while (n) {
TreeLinkNode * next = NULL; // the first node of next level
TreeLinkNode * prev = NULL; // previous node on the same level
for (; n; n=n->next) {
if (!next) next = n->left?n->left:n->right;
if (n->left) {
if (prev) prev->next = n->left;
prev = n->left;
}
if (n->right) {
if (prev) prev->next = n->right;
prev = n->right;
}
}
n = next; // turn to next level
}
}
Leetcode Populating Next Right Pointers in Each Node II
原文:http://blog.csdn.net/kenden23/article/details/18361993