摘要：This question is standard level BSF. From this question, we can see the best time to mark a node is the time when it is added to a queue.
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posted @ 2017-06-16 07:36
06 2017 档案
摘要：Note: This is to count the level. At beginning, since there are so many zombies, we need to add all of them to the Queue to start with. It is differen
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posted @ 2017-06-15 11:22
摘要：This is Matrix Problem which can be converted to BSF: Loop through each node, if it is island, find the edge. Inside the mark them all as sea. So it w
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posted @ 2017-06-15 07:46
摘要：Only Path :判断是否只存在一个拓扑排序的序列 只需要保证队列中一直最多只有1个元素即可 tricky part:1) org has all the nodes. That is why each time there should only be 1 element in the que
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posted @ 2017-06-15 06:17
摘要：Note: 1) For integer node 0~n, we don't need a hashmap, we can just use array to act like a map; 2) How to initial List[] is very important 3) We need
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posted @ 2017-06-14 08:38
摘要：Topological Sorting is very important, almost every company has a question related to it. Goal: To find an proper order making the front nodes are ind
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posted @ 2017-06-14 06:52
摘要：Note: BSF: if it needs to check the value, should check it when it is coming from Queue. The logic is most clear to do it there. This question is to a
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posted @ 2017-06-14 06:06
摘要：This question is the same as Binary Tree Treverse in Level Order. Instead of a list, we need ListNode to remeber the result. The tricky part is that w
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posted @ 2017-06-13 09:36
摘要：Similar to Binary Tree Level Order Travel, just need to add a flag to remeber the order. Change the order to opposite after the level is over.
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posted @ 2017-06-13 09:18
摘要：This is very similar to Binary Tree Level Order Traversal. Just reverse the order of the list. ArrayList has API- > add(int index, Element) or Collect
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posted @ 2017-06-13 09:06
摘要：This question is very important: 1) Serialization: a. Don't try to do all the things at once. The first step is to get a list of all the tree node. Th
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posted @ 2017-06-13 08:42
摘要：Note: This question is good summary for this kind of problem. 1) Once you get the root, loop through all the children. Get the max up/down/max from th
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posted @ 2017-06-07 08:50
摘要：Note: O(n) This question is the extension of the version one. The longest consecutive Sequence could be found in "Parent->Children" or "Children -> Pa
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posted @ 2017-06-07 07:54
摘要：/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { ...
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posted @ 2017-06-07 06:22
摘要：Note: This is question is very similar to LCA original. The only difference is that the node may not exist. So if the node is not exsit, of course the
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posted @ 2017-06-07 05:45
摘要：Note 这道题和一不同，给了一个Node向上回溯的可能，所以不需要recersive的找。因为之前的那个题不能回头，所以必须先到最下面（或者找的A和B）。这道题我们只需要把A和B的path记住就可以了，然后比较path中从root到A或者B，一直到开始不一样的时候停止，那个最后一个一样的就是LCA
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posted @ 2017-06-06 11:27
摘要：Note: // 在root为根的二叉树中找A,B的LCA: // 如果找到了就返回这个LCA // 如果只碰到A，就返回A // 如果只碰到B，就返回B // 如果都没有，就返回null Just consider relationship between left, right and root
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posted @ 2017-06-02 08:03
摘要：Note: It is easier to use divided conquer. As you can see, the question is just to add right to left's last. Here we care more about last then the top
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posted @ 2017-06-02 06:54
摘要：Node: Quick Select: Avg O(N) Thought: put all the element less(more) than pivot to the first part, the rest to the second part. And Kth only falls to
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posted @ 2017-06-02 06:09
摘要：Important: using a/b < c/d => a*d < c*b to check, we don't need to do the type transfer For the tree, if we need to get the max/min tree node, it is b
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posted @ 2017-06-01 08:29
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