[LeetCode] 113. Path Sum II Java

题目：

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For 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]
]

 题意及分析：输出从根节点到叶子节点的路径上所有点之和==sum的路径。遍历树，遍历到每个非叶子节点，减去该点的值，并加入路径中(list)，到达叶子节点时，判断，如果sum-node.val==0那么将该点添加到路径，并添加到结果中，否则继续遍历。对每个点，用一个list保存，根节点到当前点的路径。

代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        if(root==null) return res;
        List<Integer> list = new ArrayList<>();
        getPath(res,list,root,sum);
        return res;
    }

    public void getPath( List<List<Integer>> res,List<Integer> list,TreeNode node,int sum){
        List<Integer> newList = new ArrayList<>(list);
        if(node.left==null&&node.right==null){ //遍历到根节点
            if(sum-node.val==0) {   //满足条件
                newList.add(node.val);
                res.add(new ArrayList<>(newList));
                newList.clear();
            }
            return;
        }
        newList.add(node.val);
        if(node.left!=null){
            getPath(res,newList,node.left,sum-node.val);
        }
        if(node.right!=null){
            getPath(res,newList,node.right,sum-node.val);
        }
    }

}

 

 

 

 

 

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