Populating Next Right Pointers in Each Node II 题解
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题目来源:https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/description/
Description
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.
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
Solution
class Solution {
private:
void connectNode(vector<TreeLinkNode*>& v) {
int size = v.size();
for (int i = 0; i <= size - 2; i++) {
v[i] -> next = v[i + 1];
}
}
struct LevelNode {
TreeLinkNode* node;
int level;
LevelNode(TreeLinkNode* n, int l) {
node = n;
level = l;
}
};
const int InitLevel = 1;
public:
void connect(TreeLinkNode *root) {
if (root == NULL)
return;
int curLevel = InitLevel;
vector<TreeLinkNode*> v;
queue<LevelNode> q;
q.push(LevelNode(root, InitLevel));
while (!q.empty()) {
LevelNode ln = q.front();
q.pop();
if (ln.node -> left != NULL) {
q.push(LevelNode(ln.node -> left, ln.level + 1));
}
if (ln.node -> right != NULL) {
q.push(LevelNode(ln.node -> right, ln.level + 1));
}
if (ln.level != curLevel) {
connectNode(v);
v.clear();
curLevel++;
}
v.push_back(ln.node);
}
connectNode(v);
}
};
解题描述
这道题是Populating Next Right Pointers in Each Node的升级版:这道题里面的二叉树不是完全二叉树,所以每一层的节点数目没有办法提前计算。我想到的算法是,使用一个新的结构体来记录队列中的节点,然后这个结构体中包含另外一个属性——即层次编号。通过层次编号来区分不同层次的节点,在层次编号发生变化的时候对当前层次记录表中的节点进行next连接即可。