112. Path Sum

最新推荐文章于 2024-01-17 15:29:33 发布

原创最新推荐文章于 2024-01-17 15:29:33 发布 · 96 阅读

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本文探讨了在给定的二叉树中寻找一条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)算法，我们检查每个可能的路径是否满足条件。当遇到叶子节点且当前路径和等于目标值时，返回真。否则继续遍历左子树和右子树。

题目描述

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.

思路与实现

class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }

        return dfs(root, sum);
    }

    private boolean dfs(TreeNode node, int sum) {
        if (node == null) {
            return false;
        }
        if (node.val == sum && node.left == null && node.right == null) {
            return true;
        }

        return dfs(node.left, sum - node.val) || dfs(node.right, sum - node.val);
    }
}