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【LeetCode】Populating Next Right Pointers in Each Node
题意:
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
思路:
要为每个节点的 next 指针赋值,指向本层的右边节点,最右节点 next 指针为空。下面是两种解法。
代码:
C++
解一:
/** * 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) {} * }; */ #include<queue> class Solution { public: void connect(TreeLinkNode *root) { /* 定义处理队列 */ queue<TreeLinkNode*> q; if(root != NULL) q.push(root); while(!q.empty()){ /* 临时队列,用于存储下一层节点 */ queue<TreeLinkNode *> next; while(!q.empty()) { TreeLinkNode* cur = q.front(); q.pop(); /* 修改 next 指针 */ if(!q.empty()) { TreeLinkNode* val = q.front(); cur->next = val; } else cur->next = NULL; /* 添加下一层节点 */ if(cur->left != NULL) next.push(cur->left); if(cur->right != NULL) next.push(cur->right); } /* 复制队列 */ if(!next.empty()) q = next; } } };
解二:
/** * 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) { /*节点为空,直接返回*/ if (root==NULL) return; /*如果父节点有左节点和右节点,那么左节点的下一个是父节点的右节点*/ if (root->left && root->right) root->left->next = root->right; /*如果父节点有右节点和 next 指针不空,则右节点指向父节点 next 节点的左节点 */ if (root->next && root->right) root->right->next = root->next->left; /*左右孩子做连接*/ connect(root->left); connect(root->right); } };
Python:
# Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # self.next = None class Solution: # @param root, a tree node # @return nothing def connect(self, root): if root == None: return if root.left != None and root.right != None: root.left.next = root.right if root.right != None and root.next != None: root.right.next = root.next.left self.connect(root.left) self.connect(root.right)
【LeetCode】Populating Next Right Pointers in Each Node
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