path-sum Java code

最新推荐文章于 2024-01-25 20:18:16 发布

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#java #path #equals #sum

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本文介绍了一种算法，用于判断给定的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。通过递归遍历的方法实现这一目标。

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 andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

/**
 * 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;
        int now=0;
        return  digui(root,sum,now);
    }
    boolean digui(TreeNode root, int sum,int now){
         now+=root.val;
        if(sum==now&&root.left==null&&root.right==null){
            return true;
        }
        boolean left=false,right=false;

         if(root.left!=null)
             left= digui(root.left,sum,now);
        if(root.right!=null)
             right= digui(root.right,sum,now);
        return left||right;

    }
}