LC116 & 117 Populating Next Right Pointers in Each Node
LC116.Populating Next Right Pointers in Each Node
给一个完美二叉树的节点添加一个指向右边下一个节点的指针
方法1:(递归)二叉树前序遍历,当前遍历到的点,如果左孩子存在,左孩子的 next 指向右孩子,此外,如果当前节点 next != NULL ,还需设置右孩子的 next 指针。
class Solution { public: void dfs(Node* root) { if (root == NULL) return; if (root->left != NULL) { root->left->next = root->right; if (root->next) { root->right->next = root->next->left; } } dfs(root->left); dfs(root->right); } Node* connect(Node* root) { auto p = root; dfs(p); return root; } };
方法2:(迭代)二叉树的层序遍历,从根节点开始,设置好下一层的 next 指针,再从下一层开始,设置好下下一层的 next 指针,以此类推。
class Solution { public: Node* connect(Node* root) { if (root == NULL) { return root; } auto pre = root, cur = pre; while (pre->left) { cur = pre; while (cur) { cur->left->next = cur->right; if (cur->next) cur->right->next = cur->next->left; cur = cur->next; } pre = pre->left; } return root; } };
LC117. Populating Next Right Pointers in Each Node II
同上题,但是二叉树是任意的。
方法1:(递归)还是二叉树的前序遍历,但是由于当前节点的孩子节点的 next 指针依赖于当前节点的 next 指针,因此当前节点的右边的所有 next 指针必须先设置好,所以先序遍历的顺序应该为:根节点$\Rightarrow$右孩子$\Rightarrow$左孩子。
class Solution { public: void dfs(Node* root) { if (!root) return; auto p = root->next; while (p) { if (p->left) { p = p->left; break; } if (p->right) { p = p ->right; break; } p = p->next; } if (root->right) root->right->next = p; if (root->left) root->left->next = root->right ? root->right : p; dfs(root->right); dfs(root->left); } Node* connect(Node* root) { auto p = root; dfs(p); return root; } };
方法2:(迭代)还是二叉树的层序遍历,不同于第一题,遍历当前层时,需要记录下一层的第一个节点的位置,遍历完当前层后,从下一层第一个节点开始遍历。
class Solution { public: Node* connect(Node* root) { Node *head = NULL, *prev = NULL, *cur = root; while (cur) { while (cur) { if (cur->left) { if (prev) { prev->next = cur->left; } else { head = cur->left; } prev = cur->left; } if (cur->right) { if (prev) { prev->next = cur->right; } else { head = cur->right; } prev = cur->right; } cur = cur->next; } cur = head; head = NULL; prev = NULL; } return root; } };
上述方法可以用一个dummy节点,将两个循环变为一个,复杂度一样,但是代码更简洁:
class Solution { public: Node* connect(Node* root) { Node* dummyHead = new Node(0); Node* pre = dummyHead, *cur = root; while (cur) { if (cur->left) { pre->next = cur->left; pre = pre->next; } if (cur->right) { pre->next = cur->right; pre = pre->next; } cur = cur->next; if (cur == NULL) { pre = dummyHead; cur = pre->next; pre->next = NULL; } } return root; } };