【leetcode】Populating Next Right Pointers in Each Node II
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Question:
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
Anwser 1:
/** * 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 if(NULL == root){ return; } TreeLinkNode *cur = root->next; TreeLinkNode *p = NULL; while(cur != NULL){ // find last right node (left or right) if(cur->left) { p = cur->left; break; } if(cur->right){ p = cur->right; break; } cur = cur->next; } if(root->right){ root->right->next = p; } if(root->left){ root->left->next = root->right ? root->right : p; } connect(root->right); // from right to left connect(root->left); }};注意点:
1) list为非完美二叉树,右分支可能为空,因此从right -> left 遍历
2) 从最右分支开始查找,且root没有 left 节点,则找 right 节点
Anwser 2:
/** * 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 if(NULL == root){ return; } queue<TreeLinkNode *> Q; // save one line root(s) queue<TreeLinkNode *> Q2; // save next one line root(s), swap with Q Q.push(root); while(!Q.empty()){ TreeLinkNode *tmp = Q.front(); Q.pop(); if(tmp->left) Q2.push(tmp->left); if(tmp->right) Q2.push(tmp->right); if(Q.empty()){ tmp->next = NULL; queue<TreeLinkNode*> tmpQ = Q; // swap queue Q = Q2; Q2 = tmpQ; } else { tmp->next = Q.front(); } } }};注意点:
1) 新增一个Q2队列,保存下一行的全部元素,辅助判断是最后一个元素(Q为空)则置为NULL
2) queue队列实现比递归要好
Anwser 3:
/** * 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 if(NULL == root){ return; } queue<TreeLinkNode *> Q; // save one line root(s) Q.push(root); Q.push(NULL); // flag while(!Q.empty()){ TreeLinkNode *tmp = Q.front(); Q.pop(); if(tmp != NULL){ // check flag if(tmp->left) Q.push(tmp->left); if(tmp->right) Q.push(tmp->right); tmp->next = Q.front(); } else { if(Q.empty()){ // pop flag = NULL, then check is empty break; } Q.push(NULL); } } }};注意点:
对比第二种方法,改进点是每一行结束处,采用NULL作为标志位
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