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 7After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ \ 4-> 5 -> 7 -> NULLShow Similar Problems
It’s similar to the former problem Populating Next Right Pointers in Each Node I. The key to this question is to implement two functions: findNextLevelStartNode and findNextLevelNextNode
/** * 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) { TreeLinkNode * p, * startNode; TreeLinkNode * q; while((startNode = findNextLevelStartNode(root)) != NULL){ p = startNode; while((q = findNextLevelNextNode(root, p)) != NULL ){ p->next = q; p = q; } root = startNode; } } //find the leftmost node in next level //if return NULL, traversal is end TreeLinkNode * findNextLevelStartNode(TreeLinkNode * root){ if(root == NULL){ return NULL; } if(root->left != NULL) return root->left; if(root->right != NULL) return root->right; return findNextLevelStartNode(root->next); } //find next node //if return NULL, current level is end TreeLinkNode * findNextLevelNextNode(TreeLinkNode *& root, TreeLinkNode * node){ if(root == NULL){ return NULL; } if(node == NULL){ if(root->left != NULL) return root->left; if(root->right != NULL) return root->right; root = root->next; return findNextLevelNextNode(root, node); } if(node == root->left){ if(root->right != NULL) return root->right; else{ root = root->next; return findNextLevelNextNode(root, node); } } if(node == root->right){ root = root->next; return findNextLevelNextNode(root, node); } if(root->left != NULL) return root->left; if(root->right != NULL) return root->right; root = root->next; return findNextLevelNextNode(root, node); } };