Leetcode[257] Binary Tree Paths

LeetCode[257] Binary Tree Paths

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

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:
["1->2->5", "1->3"]

Recursion

复杂度
O(N), O(H)

思路
基本算法，递归。

代码

public List<String> binaryTreePaths(TreeNode root) {
    List<String> list = new LinkedList<>();
    helper(root, "", list);
    return list;
}

public void helper(TreeNode root, String cur, List<String> list) {
    if(root == null) return null;
    cur += root.val;
    if(root.left == null && root.right == null) {
        list.add(cur);
        return;
    }
    cur += "->";
    helper(root.left, cur, list);
    helper(root.right, cur, list);
}

Iteration

复杂度
O(N), O(N)

思路
递归用stack和stack进行dfs。

代码

public List<String> paths(TreeNode root) {
    StringBuilder builder = new StringBuilder();
    List<String> res = new LinkedList<>();
    if(root == null) return res;
    Stack<TreeNode> stack = new Stack<>();
    Set<TreeNode> set = new HashSet<>();
    stack.push(root);
    builder.append(root.val);
    while(!stack.isEmpty()) {
        TreeNode cur = stack.peek();
        set.add(cur);
        if(cur.left != null && !set.contains(cur.left)) {
            builder.append(cur.left.val);
            stack.push(cur.left);
            continue;
        }
        if(cur.right != null && !set.contains(cur.right)) {
            builder.append(cur.right.val);
            stack.push(cur.right);
            continue;
        }
        res.add(builder.toString());
        builder.deleteCharAt(builder.length() - 1);
    }
    return res;
}