Path Sum
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)
return sum==root.val?true:false;
return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);
}
}Path Sum II
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]
]
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
ArrayList<ArrayList<Integer>> results=new ArrayList<ArrayList<Integer>>();
if(root==null)
return results;
helper(results,new ArrayList<Integer>(),root,sum);
return results;
}
private void helper(ArrayList<ArrayList<Integer>> results,ArrayList<Integer> path,TreeNode root,int sum){
if(root==null)
return;
path.add(root.val);
if(root.left==null&&root.right==null){
if(root.val==sum){
ArrayList<Integer> temp=new ArrayList<Integer>(path);
results.add(temp);
}
}
helper(results,path,root.left,sum-root.val);
helper(results,path,root.right,sum-root.val);
path.remove(path.size()-1);
}
}