[LeetCode 112] Path Sum

문제 바로가기

Description

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Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1

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• Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
• Output: true
• Explanation: The root-to-leaf path with the target sum is shown.

Example 2

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• Input: root = [1,2,3], targetSum = 5
• Output: false
• Explanation: There are two root-to-leaf paths in the tree:
  • (1 –> 2): The sum is 3.
  • (1 –> 3): The sum is 4.
  • There is no root-to-leaf path with sum = 5.

Example 3

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• Input: root = [], targetSum = 0
• Output: false
• Explanation: Since the tree is empty, there are no root-to-leaf paths.

Constraints

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• The number of nodes in the tree is in the range [0, 5000].
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

Code

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내 제출

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        return dfs(root, targetSum);
    }

    private boolean dfs(TreeNode root, int s) {
        if (root == null) {
            return false;
        }
        s -= root.val;
        if (root.left == null && root.right == null && s == 0) {
            return true;
        }
        return dfs(root.left, s) || dfs(root.right, s);
    }
}

Runtime	Memory
0 ms	43.3 MB

다른 풀이

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {


    public boolean hasPathSum(TreeNode root, int targetSum) {
        
        if(root == null)
            return false;

        // System.out.println(targetSum - root.val);
        if(root.left==null && root.right==null) {
            
            if(targetSum - root.val == 0) {
                System.out.println("true");
                return true;
            }
        }

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

    }
}

Reference

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• https://github.com/doocs/leetcode/blob/main/solution/0100-0199/0111.Minimum%20Depth%20of%20Binary%20Tree/Solution.java