Leetcode solution 257: Binary Tree Paths

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 Problem Statement 

Input:

   1
 /   2     3
   5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Problem link

Video Tutorial

You can find the detailed video tutorial here

• Youtube
• B站

Thought Process

Easy problem. Use pre-order traversal from root to leaf and keep the recursion path. The recursion returning condition would be when a node doesn’t have left and right children. Note use a string to keep appending is easier than using a string builder, because we need to pop out and reset the string builder.

Caveat

• Handle the single node situation when no "->". E.g.,  A vs A->B

There is also an iterative implementation of this using one stack, similar to BST iterator using one stack problem. 

Solutions

Pre-order traversal

 1 private List<String> res; // stores the final output
 2 
 3     public List<String> binaryTreePaths(TreeNode root) {
 4         this.res = new ArrayList<>();
 5         helper(root, "");
 6         return this.res;
 7     }
 8 
 9     // helper function that does basic depth first traversal
10     private void helper(TreeNode root, String str) {
11         if(root == null) {
12             return;
13         }
14 
15         if(root.left==null && root.right==null) // reach a leaf node, thus completes a path and need to add it into the output
16             this.res.add(str + root.val);
17         else {
18             str += root.val + "->";
19             helper(root.left, str);
20             helper(root.right, str);
21         }
22         return;
23     }

Time Complexity: O(N), each node is visited once

Space Complexity: O(N), N is the total nodes since that‘s the string length you need

References

• Leetcode official solution

Leetcode solution 257: Binary Tree Paths

标签：class   lan   span   lin   bsp   pen   int   tail   length   

原文地址：https://www.cnblogs.com/baozitraining/p/11441305.html