LeetCode – Path Sum II (Java)
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
the method returns the following:
[ [5,4,11,2], [5,8,4,5] ]
Analysis
This problem can be converted to be a typical depth-first search problem. A recursive depth-first search algorithm usually requires a recursive method call, a reference to the final result, a temporary result, etc.
Java Solution
public List<ArrayList<Integer>> pathSum(TreeNode root, int sum) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if(root == null) return result; ArrayList<Integer> l = new ArrayList<Integer>(); l.add(root.val); dfs(root, sum-root.val, result, l); return result; } public void dfs(TreeNode t, int sum, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> l){ if(t.left==null && t.right==null && sum==0){ ArrayList<Integer> temp = new ArrayList<Integer>(); temp.addAll(l); result.add(temp); } //search path of left node if(t.left != null){ l.add(t.left.val); dfs(t.left, sum-t.left.val, result, l); l.remove(l.size()-1); } //search path of right node if(t.right!=null){ l.add(t.right.val); dfs(t.right, sum-t.right.val, result, l); l.remove(l.size()-1); } } |
<pre><code> String foo = "bar"; </code></pre>
Leave a comment
It still can be solved with bfs. By using the Path Sum solution, you will need extra list collections and a queue to store the lists while you traversing
public class Solution {
public List<List> pathSum(TreeNode root, int sum) {
List<List> result = new ArrayList<List>();
if(root == null) return result;
LinkedList nodes = new LinkedList();
LinkedList sumValues = new LinkedList();
nodes.add(root);
sumValues.add(root.val);
LinkedList<List> l = new LinkedList<List>();
List a = new ArrayList();
a.add(root.val);
l.add(a);
while(!nodes.isEmpty()){
Integer sumValue = sumValues.poll();
TreeNode t = nodes.poll();
a = l.poll();
if(t.left == null && t.right == null && sumValue == sum){
result.add(a);
}
else{
if(t.right != null) {
nodes.add(t.right);
sumValues.add(t.right.val + sumValue);
List r = new ArrayList();
r.addAll(a);
r.add(t.right.val);
l.add(r);
}
if(t.left != null) {
nodes.add(t.left);
sumValues.add(t.left.val + sumValue);
List le = new ArrayList();
le.addAll(a);
le.add(t.left.val);
l.add(le);
}
}
}
return result;
}
}
I wrote a solution but it was wrong. Someone could help?
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List> pathSum(TreeNode root, int sum) {
List<List> list = new ArrayList();
List sublist = new ArrayList();
Record record = new Record(0, sublist);
record = DFS(root, sum, record);
if(record.hasPath()) {
list.add(record.getList());
}
return list;
}
public Record DFS(TreeNode node, int sum, Record record) {
if(node == null) {
return null;
}
if(node.left == null && node.right == null) {
if(record.getVal() + node.val == sum) {
record.hasPath();
record.add(node.val);
return record;
} else {
return record;
}
}
record.hasPath();
record.add(node.val);
record.updateVal(node.val);
return (DFS(node.left, sum, record) || DFS(node.right, sum, record));
//return DFS(node.left, sum, record);
}
}
class Record {
private int calculatedVal;
private List sublist;
private boolean foundPath;
public Record(int c, List s) {
this.calculatedVal = c;
this.sublist = s;
this.foundPath = false;
}
public int getVal() {
return calculatedVal;
}
public void updateVal(int val) {
calculatedVal += val;
}
public List getList() {
return sublist;
}
public void add(int val) {
sublist.add(val);
}
public boolean hasPath() {
foundPath = true;
return foundPath;
}
}