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] ]
Approach 1:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private List<List<Integer>> resultList = new ArrayList<List<Integer>>();
public void pathSumInner(TreeNode root, int sum, Stack<Integer>path) {
path.push(root.val);
if(root.left == null && root.right == null)
if(sum == root.val) resultList.add(new ArrayList<Integer>(path));
if(root.left!=null) pathSumInner(root.left, sum-root.val, path);
if(root.right!=null)pathSumInner(root.right, sum-root.val, path);
path.pop();
}
public List<List<Integer>> pathSum(TreeNode root, int sum) {
if(root==null) return resultList;
Stack<Integer> path = new Stack<Integer>();
pathSumInner(root, sum, path);
return resultList;
}
}
Approach 2:
/**
* 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<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new ArrayList<>();
helper(root, sum, result, new ArrayList<>());
return result;
}
private void helper(TreeNode root, int sum, List<List<Integer>> result, List<Integer> list) {
if (root == null) {
return;
}
list.add(root.val);
if (root.left == null && root.right == null && root.val == sum) {
result.add(new ArrayList<>(list));
list.remove(list.size()-1);
return;
}
helper(root.left, sum - root.val, result, list);
helper(root.right, sum - root.val, result, list);
list.remove(list.size()-1);
}
}