【leetcode】112. Path Sum

最新推荐文章于 2023-11-14 10:16:47 发布

ghscarecrow

最新推荐文章于 2023-11-14 10:16:47 发布

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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，使得路径上的所有值之和等于给定的总和。通过递归算法实现，详细解析了代码逻辑与实现细节。

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.

思考：二叉树问题，可以想到的就是采用递归进行编程，由于是需要从根结点到叶子结点找到一条路径，所以采用的是从顶向下的递归方式

/**
 * 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) {
        return dfs(root,sum);
    }
    
    public boolean dfs(TreeNode root, int sum){
        //reach the node is null,mean that this path can not meet a criterion
        if(root == null){
            return false;
        }
        sum -= root.val;
        if((root.left==null) && (root.right==null) ){
            return 0 == sum;
        }
        return (dfs(root.left,sum)||dfs(root.right,sum));
    }
}