0113. Path Sum II - kumaeki/LeetCode GitHub Wiki

0113. Path Sum II

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where each path's sum equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22

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

Example 2:

Input: root = [1,2,3], targetSum = 5

Output: []

Example 3:

Input: root = [1,2], targetSum = 0

Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

解法1

在每个节点,对应的新增一个list, 来保存可能的结果.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> result = new ArrayList<>();
        helper(result, new ArrayList<Integer>(), root, 0, targetSum);
        return result;
    }
    
    private void helper(List<List<Integer>> result, List<Integer> list, TreeNode node, int sum, int target){
        if(node == null)
            return;

        sum += node.val;

        if(node.left == null && node.right == null)
            if(sum == target){
                List<Integer> temp = new ArrayList<Integer>(list);
                temp.add(node.val);
                result.add(temp);
            }else
                return;
        
        if(node.left != null){
            List<Integer> temp = new ArrayList<Integer>(list);
            temp.add(node.val);
            helper(result, temp, node.left, sum, target);
        }
            
        
        if(node.right != null){
            List<Integer> temp = new ArrayList<Integer>(list);
            temp.add(node.val);
            helper(result, temp, node.right, sum, target);
        }
            
        
    }
}

解法2

只有在符合条件的结果出现时才新建list, 其他时候只保存一个list, 在结束当前node的计算后, 从list的最后减去当前node的val.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        helper(res, new ArrayList<Integer>(), root, sum);
        return res;
    }
    
    private void helper(List<List<Integer>> res, List<Integer> list, TreeNode node, int k){
        if(node == null)
            return;
        
        int val = node.val;
        k -= val;
        list.add(node.val);
        if(node.left == null && node.right == null && k == 0){
            res.add(new ArrayList<Integer>(list));
        }else{
            helper(res, list, node.left, k);
            helper(res, list, node.right, k);
        }
        list.remove(list.size()-1);
    }
}
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