117. Populating Next Right Pointers in Each Node II

问题描述

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

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

解决思路

116其实就是117的一个特例。所以这里直接贴出117的解决办法就好了。主要是利用递归。一层一层连接，每一层都找到开始的节点，然后用这个节点递归。其实这道题主要是情况讨论题，不难。

代码

/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
*  int val;
*  TreeLinkNode *left, *right, *next;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
helper(root);
}
void helper(TreeLinkNode *root) {
if (!root)
return;
TreeLinkNode *cur = root;
TreeLinkNode *next_pre = NULL,*next_begin = NULL;

while(cur != NULL) {
if (!cur->left && !cur->right) {
cur = cur->next;
} else if (cur->left && cur->right) {
if (!next_pre) {
cur->left->next = cur->right;
next_pre = cur->right;
next_begin =cur->left;
} else {
next_pre->next = cur->left;
cur->left->next = cur->right;
next_pre = cur->right;
}
cur = cur->next;
} else if (!cur->right) {
if (!next_pre) {
next_pre = cur->left;
next_begin = cur->left;
} else {
next_pre->next = cur->left;
next_pre = cur->left;
}
cur = cur->next;
} else {
if (!next_pre) {
next_pre = cur->right;
next_begin = cur->right;
} else {
next_pre->next = cur->right;
next_pre = cur->right;
}
cur = cur->next;
}
}
helper(next_begin);
}
};