LeetCode-112：Path Sum（路径总和）

最新推荐文章于 2025-12-17 20:27:06 发布

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#算法 #二叉树

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本文探讨了如何判断一棵二叉树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数值。通过深度优先搜索算法，我们实现了在Java、Python和Scala中的解决方案。

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.

Note: A leaf is a node with no children.

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

深度优先搜索

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
   public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.right == null && root.left == null) {
            return sum - root.val == 0;
        }
        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
    }
}

Python实现

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if root is None:
            return False
        if root.left is None and root.right is None:
            return sum - root.val == 0
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)

Scala实现

/**
 * Definition for a binary tree node.
 * class TreeNode(var _value: Int) {
 *   var value: Int = _value
 *   var left: TreeNode = null
 *   var right: TreeNode = null
 * }
 */
object Solution {
    def hasPathSum(root: TreeNode, sum: Int): Boolean = {
        if (root == null) {
            return false
        }
        if (root.left == null && root.right == null) {
            return sum - root.value == 0;
        }
        return hasPathSum(root.left, sum-root.value) || hasPathSum(root.right, sum-root.value)
    }
}