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 andsum =
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 andsum = 22,
5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.
代码 1
import java.util.ArrayList; class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } public class Solution { ArrayList<ArrayList<Integer>> result=new ArrayList<ArrayList<Integer>>(); ArrayList<Integer> arr=new ArrayList<Integer>(); public boolean hasPathSum(TreeNode root, int sum) { if(root==null)return false; isPath(root,0,sum); if(result.isEmpty())return false; else return true; } private void isPath(TreeNode root, int sum, int target) { if(root==null)return; else{ sum+=root.val; arr.add(root.val); if(root.left==null&&root.right==null&&sum==target){ result.add(new ArrayList<Integer>(arr)); } isPath(root.left, sum, target); isPath(root.right, sum, target); arr.remove(arr.size()-1); sum-=root.val; } } }
代码二:
import java.util.ArrayList; class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } public class Solution { public boolean hasPathSum(TreeNode root, int sum) { return hasPathSumHelper(root, sum); } private boolean hasPathSumHelper(TreeNode root, int sum) { // TODO Auto-generated method stub if(root==null)return false; if(root.left==null&&root.right==null&&sum==root.val)return true; return hasPathSumHelper(root.left, sum-root.val)||hasPathSumHelper(root.right, sum-root.val); } }
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