Populating next right pointers in Each Node
Given a binary tree
struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; }
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL
.
Initially, all next pointers are set to NULL
.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1 / \ 2 3 / \ / \ 4 5 6 7
After calling your function, the tree should look like:
1 -> NULL / \ 2 -> 3 -> NULL / \ / \ 4->5->6->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: struct Node{ TreeLinkNode * p; int layer; Node(TreeLinkNode *p, int layer){ this->p = p; this->layer = layer; } }; void connect(TreeLinkNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function if (!root){ return; } queue<Node > q; q.push(Node(root,0)); int layer2 = 0; TreeLinkNode * temp = NULL; while(!q.empty()){ Node node = q.front(); q.pop(); if (layer2 == node.layer){ if (0 == layer2){ temp = node.p; temp->next = NULL; }else{ temp->next = node.p; temp = temp->next; } }else{ temp = node.p; temp->next = NULL; layer2++; } if (node.p->left){ q.push(Node(node.p->left,node.layer +1)); } if (node.p->right){ q.push(Node(node.p->right, node.layer + 1)); } } } };
the space complexity is not constant extra space, which is O(width). Even any binary tree, it is ok.
But, how about constant extra space, i dont know..