Js源码
113. Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ /
11 13 4
/ \ /
7 2 5 1
Return:
[ [5,4,11,2], [5,8,4,5] ] 思考路程 1 递归遍历即可
236. Lowest Common Ancestor of a Binary Tree
- Lowest Common Ancestor of a Binary Tree
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 Output: 3 Explanation: The LCA of nodes 5 and 1 is 3. Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4 Output: 5 Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Note:
All of the nodes' values will be unique. p and q are different and both values will exist in the binary tree.
思考路程 1 直接遍历即可 114. Flatten Binary Tree to Linked List Given a binary tree, flatten it to a linked list in-place.
For example, given the following tree:
1
/
2 5
/ \
3 4 6
The flattened tree should look like:
1
2
3
4
5
6
思考路程 1 很简单，之间中序遍历，即可
207. Course Schedule
There are a total of numCourses courses you have to take, labeled from 0 to numCourses-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?
Example 1:
Input: numCourses = 2, prerequisites = [[1,0]] Output: true Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. Example 2:
Input: numCourses = 2, prerequisites = [[1,0],[0,1]] Output: false Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
Constraints:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented. You may assume that there are no duplicate edges in the input prerequisites. 1 <= numCourses <= 10^5
思考路程 1， 问题可以转换为如何判断有向图中是否有环，如果没有环，则可以全部学完全部课程，如果有环，则肯定学不完 那么如何判断是否有环呢？ 1.1 本来想根据一个visited数组，把访问过的都标记，这样来不断深度搜索，最后发现，逻辑自己都被搞混了，而且好多测试用例过不去，干脆放弃 1.2 可以判断是否存在入度为0的节点，如果存在则去掉，同时更新和该节点联系的入度，如果最后发现还是存在节点，则说明存在环
心得： 逻辑如果自己设想的太复杂，或者边界太多，基本上就说明该算法是不正确的，算法算法，是首先在脑中实现，是靠脑子调试，而不是靠调试器
