【leetcode】113. Path Sum II

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum 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  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

思考:又是一道与二叉树遍历相关的题目,不难看出应使用自顶向下递归的方式来解决问题,这里还考察到了回溯法的运用。以及有一个坑那就是在往二维数组添加数组时应该新建一个数组再将新建的数组添加进去,不然,由于回溯会导致原来存进去的数组被破坏。

实现代码如下:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        if(root == null){
            return result;
        }
        List<Integer> list = new ArrayList<>();
        dfs(root,sum,list,result);
        return result;
    }
    
    public void dfs(TreeNode root,int sum,List<Integer> list,List<List<Integer>> result){
        list.add(root.val);
        sum -= root.val;
        if(root.left==null && root.right==null){
            if(sum==0){
                result.add(new ArrayList<Integer>(list));//这里要new,每添加一个数组就要创建一个新的数组
                return ;
            }
        }
        if(root.left != null){
            dfs(root.left,sum,list,result);
            //复位
            list.remove(list.size()-1);
        }
        if(root.right != null){
            dfs(root.right,sum,list,result);
            list.remove(list.size()-1);
        }
        
        
        
    }
}

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转载自blog.csdn.net/ghscarecrow/article/details/86557018

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