Path Sum

最新推荐文章于 2024-10-08 17:25:07 发布

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二叉树 (binary tree) 专栏收录该内容

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本文介绍了一种算法，用于判断给定的二叉树中是否存在从根节点到叶子节点的路径，使得路径上的所有节点值之和等于给定的目标值。

Question:

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.
Code:

/**
 * Definition for binary tree
 * 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 && sum - root.val == 0) return true;
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);        
    }
}