113. Path Sum II - Medium

发布时间：2020-12-14 05:17:57 所属栏目：大数据来源：网络整理

导读：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:

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],[5,8,5]
]

time: O(n),space: O(height)

/**
 * 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>> res = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        pathSum(root,sum,list,res);
        return res;
    }
    
    public void pathSum(TreeNode root,int sum,List<Integer> list,List<List<Integer>> res) {
        if(root == null) {
            return;
        }
        if(root.left == null && root.right == null) {
            if(sum == root.val) {
                list.add(root.val);
                res.add(new ArrayList<>(list));
                list.remove(list.size() - 1);
            }
            return;
        }
        
        list.add(root.val);
        pathSum(root.left,sum - root.val,res);
        pathSum(root.right,res);
        list.remove(list.size() - 1);
    }
}

（编辑：李大同）

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