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Populating Next Right Pointers in each Node
@H_512_14@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
广度优先搜索
思路
相当于层序遍历BST, 只是除了每层的最后一个节点其他节点都要给一个next指针指向队列里的下一个节点. 此方法I &II都适用
复杂度
时间O(n) 空间O(n)
代码
public void connect(TreeLinkNode root) {
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
if (root == null) {
return;
}
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeLinkNode cur = queue.poll();
if (i < size - 1) {//除了每层最后一个节点
TreeLinkNode next = queue.peek();
cur.next = next;
}
if (cur.left != null) {
queue.offer(cur.left);
}
if (cur.right != null) {
queue.offer(cur.right);
}
}
}递归解法
思路
node的left child 要指向right child, 如果node.next 是null的话, right child指向null, 如果不是null, right child则指向root.next的left child
复杂度
时间O(n) 空间O(1)
代码
public void connect(TreeLinkNode root) {
if (root == null) {
return;
}
if (root.left != null) {
root.left.next = root.right;
}
if (root.right != null) {
if (root.next != null) {
root.right.next = root.next.left;
} else {
root.right.next = null;
}
}
connect(root.left);
connect(root.right);
}Populating Next Right Pointers in Each Node II
递归解法
思路
这道题的binary tree不一定是完整的 所以root.right.next指向是不定的 思路就是先找到root.right 右边第一个可行node存储
注意这道题应该先遍历右边然后再左边
复杂度
时间O(n) 空间O(1)
代码
public void connect(TreeLinkNode root) {
if (root == null) {
return;
}
TreeLinkNode tem = root.next;
while (tem != null) {
if (tem.left != null) {
tem = tem.left;
break;
} else if (tem.right != null) {
tem = tem.right;
break;
} else {
tem = tem.next;
}
}
if (root.right != null) {
root.right.next = tem;
}
if (root.left != null) {
if (root.right != null) {
root.left.next = root.right;
} else {
root.left.next = tem;
}
}
connect(root.right);//此处必须先构建right然后left
connect(root.left);
}
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