[LeetCode] Populating Next Right Pointers in Each Node II

[Problem]
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

[Analysis]

层次遍历，用两个队列，Q1存储TreeLinkNode，另一个Q2存储Q1中每个结点所在的level，将同一level的TreeLinkNode连成一条线。

[Solution]

//
// 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

// null root
if(NULL == root){
return;
}

// push root node
queue<TreeLinkNode *> nodeQueue;
queue<int> levelQueue;
if(root->left != NULL){
nodeQueue.push(root->left);
levelQueue.push(1);
}
if(root->right != NULL){
nodeQueue.push(root->right);
levelQueue.push(1);
}

// lever order traversal
TreeLinkNode *pre = root;
TreeLinkNode *cur = root;
int preLevel = 0, curLevel = 0;
while(!nodeQueue.empty()){
cur = nodeQueue.front();
curLevel = levelQueue.front();

// push children
if(cur->left != NULL){
nodeQueue.push(cur->left);
levelQueue.push(curLevel + 1);
}
if(cur->right != NULL){
nodeQueue.push(cur->right);
levelQueue.push(curLevel + 1);
}

// set next
if(preLevel == curLevel){
pre->next = cur;
}
pre = cur;
preLevel = curLevel;

// pop
nodeQueue.pop();
levelQueue.pop();
}
}
};

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