[LeetCode] Populating Next Right Pointers in Each Node
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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 to NULL
.
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
解题思路:
注意这里说的是完美二叉树,即满二叉树。分层来做。记录上一层的第一个节点,然后遍历上一层节点U和下一层节点L,将U的左孩子节点指向U的右孩子节点,L的右孩子节点指向U的下一个节点(如果存在的话)的左孩子节点。
/** * 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) { if(root==NULL){ return; } TreeLinkNode* upLayerFirstNode = root; while(upLayerFirstNode->left!=NULL){ TreeLinkNode* upLayerNode = upLayerFirstNode; TreeLinkNode* node = upLayerFirstNode->left; while(true){ node->next = upLayerNode->right; node = node->next; upLayerNode = upLayerNode->next; if(upLayerNode==NULL){ break; } node->next = upLayerNode->left; node = node->next; } upLayerFirstNode = upLayerFirstNode->left; } }};
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