Yosgi Yosgi

Time: 80 minutes The first reaction when I got the question was of course to solve it with DFS. But this is a complete binary tree, so we should use the properties of a complete binary tree. You can …

Time: 120 minutes The initial idea was to use BFS, and remove a whole queue when removing it (later I learned that this is called breadth traversal) Then the double pointer finds the left and right …

Time: 60 minutes The idea is still very simple. First use DFS to add a parent to each node, and find the required target by the way. Then starting from the target, use DFS to find a point K away from …

Time: 10 minutes Just use the increasing property of in-order traversal of the binary search tree In-order traversal to find the left child Does pre exist? If not, assign , find the right child, and …

Time: 60 minutes The idea is still in-order traversal, using pre to cache the previous node. Because the in-order traversal is incremental, you must find the larger one first and then the smaller one. …

Time: 30 minutes The question is divided into two parts to be solved, search and delete The search part is to find the node. You can find the required node according to the properties of the binary …

#669. Pruning a binary search tree Release Date: March 11, 2021 Time: 20 minutes Similar to the previous question, most of them are divided into situations where the parent node is in the interval and …

Time: Not done The main reason I didn’t make it was that I kept thinking about how to use a stack. When I looked at the answer to the stack later, I found that recursion was the simplest. recursion …

Time: 25 minutes The main idea is to use the stack. The difference from the ordinary level-order traversal is that the stack is cleared every time var connect = function(root) { if (!root) return null …

Time: I copied the answers and finished it. It is easy to see that the idea is to use recursion, but the coding is the real problem. What we need to return is the root node, but the recursion is from …