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.
Solution:
/** * 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) { return hasPathSum(root, sum, 0); } public boolean hasPathSum(TreeNode root, int sum, int tempSum) { if(root == null) return false; if( root.left==null && root.right == null) return tempSum + root.val == sum; return hasPathSum(root.left, sum, tempSum+root.val) || hasPathSum(root.right, sum, tempSum+root.val); } }
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