Question

2689. Extract Kth Character From The Rope Tree

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Description

You are given the root of a binary tree and an integer k. Besides the left and right children, every node of this tree has two other properties, a string node.val containing only lowercase English letters (possibly empty) and a non-negative integer node.len. There are two types of nodes in this tree:

• Leaf: These nodes have no children, node.len = 0, and node.val is some non-empty string.
• Internal: These nodes have at least one child (also at most two children), node.len > 0, and node.val is an empty string.

The tree described above is called a _Rope_ binary tree. Now we define S[node] recursively as follows:

• If node is some leaf node, S[node] = node.val,
• Otherwise if node is some internal node, S[node] = concat(S[node.left], S[node.right]) and S[node].length = node.len.

Return_ k-th character of the string_ S[root].

Note: If s and p are two strings, concat(s, p) is a string obtained by concatenating p to s. For example, concat("ab", "zz") = "abzz".

 

Example 1:

Input: root = [10,4,"abcpoe","g","rta"], k = 6
Output: "b"
Explanation: In the picture below, we put an integer on internal nodes that represents node.len, and a string on leaf nodes that represents node.val.
You can see that S[root] = concat(concat("g", "rta"), "abcpoe") = "grtaabcpoe". So S[root][5], which represents 6th character of it, is equal to "b".

Example 2:

Input: root = [12,6,6,"abc","efg","hij","klm"], k = 3
Output: "c"
Explanation: In the picture below, we put an integer on internal nodes that represents node.len, and a string on leaf nodes that represents node.val.
You can see that S[root] = concat(concat("abc", "efg"), concat("hij", "klm")) = "abcefghijklm". So S[root][2], which represents the 3rd character of it, is equal to "c".

Example 3:

Input: root = ["ropetree"], k = 8
Output: "e"
Explanation: In the picture below, we put an integer on internal nodes that represents node.len, and a string on leaf nodes that represents node.val.
You can see that S[root] = "ropetree". So S[root][7], which represents 8th character of it, is equal to "e".

 

Constraints:

• The number of nodes in the tree is in the range [1, 103]
• node.val contains only lowercase English letters
• 0 <= node.val.length <= 50
• 0 <= node.len <= 104
• for leaf nodes, node.len = 0 and node.val is non-empty
• for internal nodes, node.len > 0 and node.val is empty
• 1 <= k <= S[root].length

Solutions

Solution 1

# Definition for a rope tree node.
# class RopeTreeNode(object):
#   def __init__(self, len=0, val="", left=None, right=None):
#     self.len = len
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def getKthCharacter(self, root: Optional[object], k: int) -> str:
    def dfs(root):
      if root is None:
        return ""
      if root.len == 0:
        return root.val
      return dfs(root.left) + dfs(root.right)

    return dfs(root)[k - 1]

/**
 * Definition for a rope tree node.
 * class RopeTreeNode {
 *   int len;
 *   String val;
 *   RopeTreeNode left;
 *   RopeTreeNode right;
 *   RopeTreeNode() {}
 *   RopeTreeNode(String val) {
 *     this.len = 0;
 *     this.val = val;
 *   }
 *   RopeTreeNode(int len) {
 *     this.len = len;
 *     this.val = "";
 *   }
 *   RopeTreeNode(int len, TreeNode left, TreeNode right) {
 *     this.len = len;
 *     this.val = "";
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public char getKthCharacter(RopeTreeNode root, int k) {
    return dfs(root).charAt(k - 1);
  }

  private String dfs(RopeTreeNode root) {
    if (root == null) {
      return "";
    }
    if (root.val.length() > 0) {
      return root.val;
    }
    String left = dfs(root.left);
    String right = dfs(root.right);
    return left + right;
  }
}

/**
 * Definition for a rope tree node.
 * struct RopeTreeNode {
 *   int len;
 *   string val;
 *   RopeTreeNode *left;
 *   RopeTreeNode *right;
 *   RopeTreeNode() : len(0), val(""), left(nullptr), right(nullptr) {}
 *   RopeTreeNode(string s) : len(0), val(std::move(s)), left(nullptr), right(nullptr) {}
 *   RopeTreeNode(int x) : len(x), val(""), left(nullptr), right(nullptr) {}
 *   RopeTreeNode(int x, RopeTreeNode *left, RopeTreeNode *right) : len(x), val(""), left(left), right(right) {}
 * };
 */
class Solution {
public:
  char getKthCharacter(RopeTreeNode* root, int k) {
    function<string(RopeTreeNode * root)> dfs = [&](RopeTreeNode* root) -> string {
      if (root == nullptr) {
        return "";
      }
      if (root->len == 0) {
        return root->val;
      }
      string left = dfs(root->left);
      string right = dfs(root->right);
      return left + right;
    };
    return dfs(root)[k - 1];
  }
};

/**
 * Definition for a rope tree node.
 * type RopeTreeNode struct {
 *    len   int
 *    val   string
 *    left  *RopeTreeNode
 *    right *RopeTreeNode
 * }
 */
func getKthCharacter(root *RopeTreeNode, k int) byte {
  var dfs func(root *RopeTreeNode) string
  dfs = func(root *RopeTreeNode) string {
    if root == nil {
      return ""
    }
    if root.len == 0 {
      return root.val
    }
    left, right := dfs(root.left), dfs(root.right)
    return left + right
  }
  return dfs(root)[k-1]
}