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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
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Initially, all next pointers are set to NULL
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Note:
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
思路:前序的递归思路。时间复杂度O(n),,空间复杂度O(1)
1 /** 2 * Definition for binary tree with next pointer. 3 * struct TreeLinkNode { 4 * int val; 5 * TreeLinkNode *left, *right, *next; 6 * TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 void connect(TreeLinkNode *root, TreeLinkNode *sibling) { 12 //处理当前节点 13 if (root == NULL) 14 return; 15 else 16 root->next = sibling; 17 18 //处理左子节点 19 connect(root->left, root->right); 20 21 //处理右子节点 22 if (sibling) { 23 connect(root->right, sibling->left); 24 } else { 25 connect(root->right, NULL); 26 } 27 } 28 void connect(TreeLinkNode *root) { 29 connect(root, NULL); 30 } 31 };
[LeetCode] Populating Next Right Pointers in Each Node
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原文地址:http://www.cnblogs.com/vincently/p/4231719.html