[Binary Tree] Path Sum II

路径总和II

这道题目10分钟就可以完成了,I've gotten a lot out of practice lately,加油

路径总和,Divide into left and right subtrees,要注意的是JavaThe path needs to be additionally saved here,Otherwise, save directlypath都是null

/** * 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 {
    
    private List<List<Integer>> ret = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
    
        if(root == null) return ret;
        List<Integer> path = new ArrayList<>();
        searchPath(root, targetSum, path);
        return ret;
    }
    private void searchPath(TreeNode root, int targetSum, List<Integer> path){
    
        
        if(root == null) return;
        if(root.left == null && root.right == null){
    
            if(root.val == targetSum){
    
                List<Integer> ans = new ArrayList<>();
                for(int i = 0; i < path.size(); i++){
      
                    ans.add(path.get(i));
                }
                ans.add(root.val);
                ret.add(ans);
            }
        }
        path.add(root.val);
        searchPath(root.left, targetSum-root.val, path);
        searchPath(root.right, targetSum-root.val, path);
        path.remove(path.size()-1);  
    }

}