leetcode oj java Path Sum

来源：互联网发布：流程优化再造的意思编辑：程序博客网时间：2024/06/12 23:53

问题描述：

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.

解决方案：从根节点依次扫描，每次减去父亲节点的值，直到直到叶子节点且减去之后值为0 即可。

package leetcode;

public class hasPathSum {

   /**
   * @param args
   */

    public static boolean hasPathSum(TreeNode root, int sum){
       if(root == null){
           return false;
       }
       if(root.right == null && root.left == null){
           if(sum - root.val == 0)
               {return true;}
       }

       return hasPathSum(root.left , sum-root.val) ||
               hasPathSum(root.right, sum-root.val);
   }

    public static void main(String[] args) {
       // TODO Auto-generated method stub
       TreeNode root = new TreeNode(5);
       TreeNode rl = new TreeNode(4);
       TreeNode rr = new TreeNode(8);
       TreeNode rll = new TreeNode(11);
       TreeNode rlll = new TreeNode(7);
       TreeNode rllr = new TreeNode(2);
       TreeNode rrl = new TreeNode(13);
       TreeNode rrr = new TreeNode(4);
       TreeNode rrrr = new TreeNode(1);

       root.left = rl;
       root.right = rl;
       rl.left = rll;
       rll.left = rlll;
       rll.right = rllr;
       rr.left = rrl;
       rr.right = rrr;
       rrr.right = rrrr;


       boolean flag = hasPathSum(root, 22);
       System.out.println(flag);

   }

}

给出的是完整的代码，可以自己调试。

提交时只需要 hasPathSum 这个方法。

运行结果：

114 / 114 test cases passed.

Status:

Accepted

Runtime: 1 ms

Submitted: 0 minutes ago

0 0