Leetcode: Path Sum

最新推荐文章于 2021-05-11 15:07:28 发布

转载最新推荐文章于 2021-05-11 15:07:28 发布 · 36 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/EdwardLiu/p/3708902.html

本文介绍了一种算法，用于判断二叉树中是否存在从根节点到叶子节点的路径，使得这条路径上所有节点值之和等于给定的数值。通过递归方式实现，提供了两种简洁高效的代码实现。

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.

比较通俗易懂一点的看法：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean hasPathSum(TreeNode root, int sum) {
12         if (root == null) return false;
13         return helper(root, sum);
14     }
15     
16     public boolean helper(TreeNode root, int remain) {
17         if (root.left==null && root.right==null) {
18             if (remain-root.val == 0) return true;
19             else return false;
20         }
21         boolean left=false, right=false;
22         if (root.left != null) {
23             left = helper(root.left, remain-root.val);
24         }
25         if (root.right != null) {
26             right = helper(root.right, remain-root.val);
27         }
28         return left || right;
29     }
30 }

精炼简洁的做法，但是不容易写：

1 public boolean hasPathSum(TreeNode root, int sum) {
2     if(root == null)
3         return false;
4     if(root.left == null && root.right==null && root.val==sum)
5         return true;
6     return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
7 }

转载于:https://www.cnblogs.com/EdwardLiu/p/3708902.html