wayetan
12/30/2013 - 6:48 AM

## Path Sum

Path Sum

``````/**
* 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]
]
*/

public class Solution {
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root != null){
ArrayList<Integer> path = new ArrayList<Integer>();
pathSumHelper(root, sum, res, path);
}
return res;
}

public void pathSumHelper(TreeNode root, int sum, ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path){
if(root == null)
return;
if(root.left == null && root.right == null && root.val == sum){
// path found.
ArrayList<Integer> apath = new ArrayList<Integer>(path);
}else{
// walk into next level
if(root.left != null)
pathSumHelper(root.left, sum - root.val, res, path);
if(root.right != null)
pathSumHelper(root.right, sum - root.val, res, path);
// recover the global variable. backtracking.
path.remove(path.size() - 1);
}
}
}``````
``````/**
* Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
* For example:
* Given the below binary tree and sum = 22,
*            5
/ \
4   8
/   / \
11  13  4
/  \      \
7    2      1
* return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
*/

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null)
return false;
if(root.left == null && root.right == null && root.val == sum)
return true;
else
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}``````