如果您对在TD的jqueryTreeTraversalprev()和在html中,td标签的()属性用于创建跨多个行的单元格感兴趣,那么这篇文章一定是您不可错过的。我们将详细讲解在TD的jqueryT
如果您对在TD的jquery Tree Traversal prev()和在html中,td标签的( )属性用于创建跨多个行的单元格感兴趣,那么这篇文章一定是您不可错过的。我们将详细讲解在TD的jquery Tree Traversal prev()的各种细节,并对在html中,td标签的( )属性用于创建跨多个行的单元格进行深入的分析,此外还有关于#Leetcode# 144. Binary Tree Preorder Traversal、102 Binary Tree Level Order Traversal、102. Binary Tree Level Order Traversal、103. Binary Tree Zigzag Level Order Traversal的实用技巧。
本文目录一览:- 在TD的jquery Tree Traversal prev()(在html中,td标签的( )属性用于创建跨多个行的单元格)
- #Leetcode# 144. Binary Tree Preorder Traversal
- 102 Binary Tree Level Order Traversal
- 102. Binary Tree Level Order Traversal
- 103. Binary Tree Zigzag Level Order Traversal
在TD的jquery Tree Traversal prev()(在html中,td标签的( )属性用于创建跨多个行的单元格)
<table> <tr> <td><input type="checkBox"></td> <td><label>Text 1</label></td> </tr> <tr> <td><input type="checkBox"></td> <td><label>Text 2</label></td> </tr> </table> <script> $('.resultchecklabel').live('click',function() { if ($(this).prev('td input:checked').val() != null) { $(this).prev('td').prev("input").removeAttr("checked"); } else { $(this).prev('td').prev("input").attr("checked","checked"); } }); </script>
谁能帮我解决这个问题?
非常感谢!
解决方法
你的代码不起作用的原因可能是因为你的上下文中的这个是label元素,而td是彼此的兄弟,而不是标签.我认为如果您有更复杂的要求 – 例如额外的造型等,这将完成您想要的并且就足够了.
var cb = $(this).closest('td') // containing TD .prev('td') // prevIoUs TD .find('input'); // input within TD if (cb.find(':checked').length == 0) { // not checked cb.attr('checked','checked'); } else { cb.removeAttr('checked'); }
#Leetcode# 144. Binary Tree Preorder Traversal
https://leetcode.com/problems/binary-tree-preorder-traversal/
Given a binary tree,return the preorder traversal of its nodes‘ values.
Example:
Input: 1 2 / 3 Output: [1,null,2,3][1,3]
Follow up: Recursive solution is trivial,Could you do it iteratively?
代码:
/** * DeFinition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x),left(NULL),right(NULL) {} * }; */ class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; if(!root) return ans; pre(ans,root); return ans; } void pre(vector<int>& ans,TreeNode* root) { ans.push_back(root -> val); if(root -> left) pre(ans,root -> left); if(root -> right) pre(ans,root -> right); } };
今天刷完了《妻子的浪漫旅行》 可以安心学习啦
102 Binary Tree Level Order Traversal
Given a binary tree,return the level order traversal of its nodes‘ values. (ie,from left to right,level by level). For example:?Given binary tree [3,9,20,null,15,7],? 3 / 9 20 / 15 7 return its level order traversal as:? [ [3],[9,20],[15,7] ] class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); if(root == null) return result; Queue<TreeNode> queue = new LinkedList<>(); // queue,stack interface?? queue.offer(root); while(!queue.isEmpty()){ int size = queue.size(); List<Integer> list = new ArrayList<>(); // create a new list here before the size for(int i = 0; i < size; i++){ TreeNode cur = queue.poll(); list.add(cur.val); if(cur.left != null) queue.offer(cur.left); if(cur.right != null) queue.offer(cur.right); } result.add(list); } return result; } } // stack,push,pop // queue. Offer,poll
102. Binary Tree Level Order Traversal
身体不好
1 class Solution { 2 List<List<Integer>> res = new ArrayList<>(); 3 public List<List<Integer>> levelOrder(TreeNode root) { 4 bfs(root,0); 5 return res; 6 7 } 8 9 public void bfs(TreeNode root,int n) { 10 if(root == null) return; 11 if(n > res.size()) { 12 res.add(new ArrayList<Integer>()); 13 } 14 res.get(n).add(root.val); 15 16 17 bfs(root.left,n+1); 18 bfs(root.right,n+1); 19 20 } 21 22 }
103. Binary Tree Zigzag Level Order Traversal
树的蜿蜒型层次遍历
要求第一层从左向右,第二层从右向左...依次类推
和102题无本质区别,加一个翻转指示器即可
# DeFinition for a binary tree node. # class TreeNode: # def __init__(self,x): # self.val = x # self.left = None # self.right = None class Solution: def zigzagLevelOrder(self,root: TreeNode) -> List[List[int]]: if not root: return [] nodelist = [root] result = [] order = True while nodelist: tmplist = [] vallist = [] for node in nodelist: vallist.append(node.val) if node.left: tmplist.append(node.left) if node.right: tmplist.append(node.right) if not order: vallist.reverse() order = not order result.append(vallist) nodelist = tmplist return result
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