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This article is about how LeetCode finds all the elements in two binary search trees. The editor thinks it is very practical, so share it with you as a reference and follow the editor to have a look.

One, two binary search trees for all the elements

1. Brief introduction of the problem.

Here are two binary search trees, root1 and root2.

Please return a list of all the integers in the "two trees" and sort them in ascending order.

2, example description

Input: root1 = [2jue 1pr 4], root2 = [1je 0pr 3]

Output: [0meme 1 pyrrine 1 pyrrine 2pime 3 pyrrine 4]

Example 2:

Input: root1 = [0jimi 10jue 10], root2 = [5meme 1je 7je 0je 2]

Output: [- 10pc0re0p0pl 1pyrrine 25pyrm 7pc10]

Example 3:

Input: root1 = [], root2 = [5Jing 1Jing 7JI 0jue 2]

Output: [0pyrrine 1 pyrrine 2pyrrine 5pr 7]

Example 4:

Input: root1 = [0mai Mei 10j 10], root2 = []

Output: [- 1010 01010]

Tip:

Each tree has a maximum of 5000 nodes.

The value of each node is between [- 10 ^ 5, 10 ^ 5].

3, the train of thought of solving the problem

Depth first search, array sort operation

4, problem solving procedure

Import java.util.ArrayList

Import java.util.Collections

Import java.util.List

Public class GetAllElementsTest3 {

Public static void main (String [] args) {

TreeNode T1 = new TreeNode (2)

TreeNode T12 = new TreeNode (1)

TreeNode T13 = new TreeNode (4)

TreeNode T2 = new TreeNode (1)

TreeNode T21 = new TreeNode (0)

TreeNode T22 = new TreeNode (3)

T1.left = T12

T1.right = T13

T2.left = T21

T2.right = T22

GetAllElements (T1, T2)

}

Public static List getAllElements (TreeNode root1, TreeNode root2) {

List list = new ArrayList ()

If (root1 = = null & & root2 = = null) {

Return list

}

List root1List = new ArrayList ()

List root2List = new ArrayList ()

DfsRoot1 (root1, root1List)

DfsRoot1 (root2, root2List)

Root1List.addAll (root2List)

Collections.sort (root1List)

Return root1List

}

Private static void dfsRoot1 (TreeNode root1, List root1List) {

If (root1 = = null) {

Return

}

If (root1.left! = null) {

DfsRoot1 (root1.left, root1List)

}

Root1List.add (root1.val)

If (root1.right! = null) {

DfsRoot1 (root1.right, root1List)

}

}

}

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