Populating Next Right Pointers in Each Node II
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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
/** * 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 findNextForLeft(TreeLinkNode *root){ if(!root || !root->left) return; TreeLinkNode *cur = root; TreeLinkNode *target = NULL; if(root -> right){ target = root -> right; }else{ while(cur -> next){ cur = cur -> next; if(cur -> left){ target = cur-> left; break; }else if(cur -> right){ target = cur -> right; break; } } } root -> left -> next = target; } void findNextForRight(TreeLinkNode *root){ if(!root || !root->right) return; TreeLinkNode *cur = root; TreeLinkNode *target = NULL; while(cur -> next){ cur = cur -> next; if(cur -> left){ target = cur-> left; break; }else if(cur -> right){ target = cur -> right; break; } } root -> right -> next = target; } void findNext(TreeLinkNode *root){ findNextForLeft(root); findNextForRight(root); } void connect(TreeLinkNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if(!root) return; root -> next = NULL; queue<TreeLinkNode *> q; q.push(root); TreeLinkNode *cur; while(!q.empty()){ cur = q.front(); q.pop(); if(cur -> left) q.push(cur -> left); if(cur -> right) q.push(cur -> right); findNext(cur); } }};
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