#112 Path Sum

最新推荐文章于 2024-10-08 17:25:07 发布

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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目标值。通过递归深度优先搜索（DFS）的方法，实现了简洁有效的解决方案。

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.

题解：

这是我开始的解法：

/**
 * 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;
        if(root.left==null && root.right==null)
            return sum==root.val;
        return hasPathSum(root.left,sum-root.val) || hasPashSum(root.right,sum-root.val);
    }
}

Submission Result: Compile ErrorMore Details 

Line 16: error: cannot find symbol: method hasPashSum(TreeNode,int)

好2，肯定要写在一个函数里的啊，怎么想的。

正解：

/**
 * 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) {
        return dfsPath(root,sum);
    }
    
    public boolean dfsPath(TreeNode root, int sum){
        if(root==null)
            return false;
        if(root.left==null && root.right==null)
            return sum==root.val;
        return dfsPath(root.left,sum-root.val) || dfsPath(root.right,sum-root.val);
    }
}