LeetCode(35)-Path Sum

最新推荐文章于 2025-12-19 16:34:09 发布

转载最新推荐文章于 2025-12-19 16:34:09 发布 · 132 阅读

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原文链接：https://my.oschina.net/fengsehng/blog/784473

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#数据结构与算法

本文探讨了如何确定一棵二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的整数。通过递归方法实现，详细解释了算法流程，并附带具体示例。

题目：

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.

思路：

题意是要从根几点root到叶子节点的所有组合，是不是等于给定的整数sum
考虑用递归，把对当前的判断，转化为root.left和sum -root.val的比较以及toot.right和sum-root.val的比较 -

代码：

/** * Definition for a binary tree node. * 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;
        }
        int count = 0;
        if(root != null){
            count = root.val;
        }
            if(root.left != null && root.right != null){
                return hasPathSum(root.left,sum-count) || hasPathSum(root.right,sum-count);
            }
            if(root.right != null){
                return hasPathSum(root.right,sum-count);
            }
            if(root.left != null){
                return hasPathSum(root.left,sum-count);
            }
            if(count == sum){
                return true;
            }
            return false;
    }
}

转载于:https://my.oschina.net/fengsehng/blog/784473