leetcode题记——Path Sum

最新推荐文章于 2023-04-28 22:25:46 发布

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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目标值。通过深度优先搜索(DFS)递归算法实现解决方案。

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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.

/**
 * 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;
        return DFS(sum,0,root);
    }
    public boolean DFS(int t , int sum , TreeNode root){
        if(root == null )
            return false;
        sum+=root.val;
        if(root.left==null && root.right==null){
            if(sum == t)
                return true;
            else
                return false;
        }
        boolean l = DFS(t,sum,root.left);
        boolean r = DFS(t,sum,root.right);
        return l||r;
    }
}