257.Binary-Tree-Paths

257. Binary Tree Paths

题目地址

https://leetcode.com/problems/binary-tree-paths/

题目描述

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

代码

Approach #1 Recursion

Time O(N) Space O(N)

/**
 * 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) {
        LinkedList<String> paths = new LinkedList();
    dfs(root, "", paths);
    return paths;
  }

  private void dfs(TreeNode root, String path, LinkedList<String> paths) {
    if (root != null) {
      path += Integer.toString(root.val);
      if (root.left == null && root.right == null) {
        paths.add(path);   // stop dfs
      } else {
        path += "->";
        dfs(root.left, path, paths);
        dfs(root.right, path, paths);
      }
    } else {
      // stop dfs
    }
  }
}

Approach #2 DFS Iteration + Two Stack

class Solution {
  public List<String> binaryTreePaths(TreeNode root) {
    LinkedList<String> paths = new LinkedList();
    if (root == null)  return paths;

    LinkedList<TreeNode> node_stack = new LinkedList();
    LinkedList<String> path_stack = new LinkedList();
    node_stack.add(root);
    path_stack.add(Integer.toString(root.val));
    TreeNode node;
    String path;
    while ( !node_stack.isEmpty() ) {
      node = node_stack.pollLast();
      path = path_stack.pollLast();
      if ((node.left == null) && (node.right == null))
        paths.add(path);
      if (node.left != null) {
        node_stack.add(node.left);
        path_stack.add(path + "->" + Integer.toString(node.left.val));
      }
      if (node.right != null) {
        node_stack.add(node.right);
        path_stack.add(path + "->" + Integer.toString(node.right.val));
      }
    }
    return paths;
  }
}