LeetCode Populating Next Right Pointers in Each Node
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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
.
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:
本题树为 完美树, 每次保存每一层的第一个节点, 执行到下一层时,上一层已经连接好了。 使用递归的话,代码更少。
非递归:
/** * 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;stack<TreeLinkNode *> MyStack;MyStack.push(root);while (!MyStack.empty()) {TreeLinkNode *head = MyStack.top();MyStack.pop();if (head->left != NULL)MyStack.push(head->left);while (head != NULL) {if (head->left == NULL)break;head->left->next = head->right;if (head->next != NULL)head->right->next = head->next->left;head = head->next;}}}};
递归:
class Solution {public:void connect(TreeLinkNode *root) {if (root == NULL)return;if (root->left != NULL) {root->left->next = root->right;root->right->next = root->next ? root->next->left : NULL;}connect(root->left);connect(root->right);}};
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