"Leetcode-Interview algorithm classic-java Implementation" "113-path Sum II (path and)"

"113-path Sum II (Path and II)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

Given a binary tree and a sum, find all root-to-leaf paths where each path ' s sum equals the Given sum.
For 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]]

Main Topic

Given a binary tree and a sum, judging from the root node of the tree to all nodes of the leaf node and whether equal to the given sum, if equal to record this path.

Thinking of solving problems

The tree is traversed, and the backtracking method is used to solve it.

Code Implementation

Tree Node class

publicclass TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) { val = x; }}

Algorithm implementation class

ImportJava.Util.ArrayList;ImportJava.Util.LinkedList;ImportJava.Util.List; PublicClass Solution {Private List<List<Integer>>ResultPrivate List<Integer>LPrivateIntsum;Privateint Cursum= 0; Public List<List<Integer>>Pathsum (TreeNode root, intsum) {Result= NewLinkedList<>();if(Root!= NULL) {This.sum = sum; L= NewLinkedList<>();        Pathsum (root); }returnResult }Private voidPathsum (TreeNode root) {if(Root!= NULL) {L.Add (Root.Val); Cursum+=Root.Valif(Root.Left== NULL &&Root.Right== NULL &&Cursum== sum) {List<Integer> List = NewLinkedList<>(); for (IntegerI:L) {List.Add (i); } result.AddList); }if(Root.Left!= NULL) {Pathsum (root.left); }if(Root.Right!= NULL) {Pathsum (root.right); } cursum-=Root.Val L.Remove (l.Size ()- 1);//Delete last}    }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java Implementation" "113-path Sum II (path and)"