leetcode 257 Binary Tree Paths

﹏ヽ暗。殇╰゛Y 2022-05-04 18:28 133阅读 0赞

Given a binary tree, return all root-to-leaf paths.

**Note:** A leaf is a node with no children.

**Example:**

Input:
    
       1
     /   \
    2     3
     \
      5
    
    Output: ["1->2->5", "1->3"]
    
    Explanation: All root-to-leaf paths are: 1->2->5, 1->3

递归方法：

1、递归终止条件，root是否为空，及是否为叶子结点

2、递归过程

/**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<String> binaryTreePaths(TreeNode root) {
            List<String> res = new ArrayList<String>();
            
            if(root == null)
                return res;
            
            if(root.left == null && root.right == null){
                res.add(root.val + "");
                return res;
            }
            
            List<String> leftS = binaryTreePaths(root.left);
            for(int i=0;i<leftS.size();i++){
                res.add(root.val + "->" + leftS.get(i));
            }
            List<String> rightS = binaryTreePaths(root.right);
            for(int i=0;i<rightS.size();i++){
                res.add(root.val + "->" + rightS.get(i));
            }
            
            return res;
        }
    }

submission中 深度优先遍历DFS：

/**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        
        public List<String> binaryTreePaths(TreeNode root) {
            List<String> result = new LinkedList<>();
            dfs(result, root, "");
            return result;
        }
        
        private void dfs(List<String> result, TreeNode node, String s) {
            if (node == null) return;
            String built = s.length() == 0 ? s + node.val : s + "->" + node.val;
            if (node.left == null && node.right == null) {
                result.add(built);
            } else {
                dfs(result, node.left, built);
                dfs(result, node.right, built);
            }
        }
    
    }