leetcode:Populating Next Right Pointers in Each Node(常数空间,连接二叉树每一层所有节点)【面试算法题】
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题目:
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
题意把给定的一个完全二叉树每一层所有的节点连接起来,只能用常数的额外空间。
用递归的做法,把当前root的左节点和它的所有深度下的右节点 与 右节点和它的所有深度下的左节点 连接。
目的就是把因为当前root左右节点而分开的子节点全部连接起来。
画个图就很容易发现,这样做就把需要连接的节点在他们的公共祖先root的递归层中一起处理。
/** * 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)return; if(root->left) { TreeLinkNode *l=root->left,*r=root->right; while(l!=NULL) { l->next=r; l=l->right; r=r->left; } connect(root->left); connect(root->right); } }};// blog.csdn.net/havenoidea
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