＃113 Path Sum II

Description

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Examples

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]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

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

思路

我纠结的点在于用什么作为中间存储介质，后面建立了一个List<List<TreeNode>>
但是看到submission里面的解法之后，确实是还有改进的空间，下次重新试试

代码

/** * 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 {
    
    List<List<TreeNode>> all = new ArrayList<>();
    List<List<Integer>> answer = new ArrayList<>();
    
    public void addAnswer(int i) {
    
        List<TreeNode> nodes = all.get(i);
        List<Integer> sum = new ArrayList<>();
        for (TreeNode node: nodes) {
    
            sum.add(node.val);
        }
        answer.add(sum);
    }
    
    public void getSum(List<Integer> sum, int targetSum) {
    
        if (all.size() == 0)
            return;
        
        List<List<TreeNode>> next = new ArrayList<>();
        List<Integer> nextSum = new ArrayList<>();
        
        for (int i = 0; i < all.size(); i++) {
    
            List<TreeNode> nodes = all.get(i);
            TreeNode node = nodes.get(nodes.size() - 1);
            int currSum = sum.get(i);
            
            if (node.left == null && node.right == null) {
    
                if (currSum == targetSum) {
    
                    addAnswer(i);
                }
                continue;
            }
            if (node.left != null) {
    
                List<TreeNode> newTmp = new ArrayList<>(nodes);
                newTmp.add(node.left);
                next.add(newTmp);
                nextSum.add(currSum + node.left.val);
            }
            if (node.right != null) {
    
                List<TreeNode> newTmp = new ArrayList<>(nodes);
                newTmp.add(node.right);
                next.add(newTmp);
                nextSum.add(currSum + node.right.val);
            }
        }
        
        all = next;
        getSum(nextSum, targetSum);
    }
    
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
    
        if (root == null)
            return new ArrayList<>();
        
        List<TreeNode> nodes = new ArrayList<>();
        nodes.add(root);
        all.add(nodes);
        List<Integer> sum = new ArrayList<>();
        sum.add(root.val);
        
        getSum(sum, targetSum);
        return answer;
    }
}