Question

538. Convert BST to Greater Tree

中文文档

Description

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.

As a reminder, a _binary search tree_ is a tree that satisfies these constraints:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

 

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -104 <= Node.val <= 104
• All the values in the tree are unique.
• root is guaranteed to be a valid binary search tree.

 

Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def convertBST(self, root: TreeNode) -> TreeNode:
    def dfs(root):
      nonlocal s
      if root is None:
        return
      dfs(root.right)
      s += root.val
      root.val = s
      dfs(root.left)

    s = 0
    dfs(root)
    return root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private int s;

  public TreeNode convertBST(TreeNode root) {
    dfs(root);
    return root;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    dfs(root.right);
    s += root.val;
    root.val = s;
    dfs(root.left);
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  int s = 0;

  TreeNode* convertBST(TreeNode* root) {
    dfs(root);
    return root;
  }

  void dfs(TreeNode* root) {
    if (!root) return;
    dfs(root->right);
    s += root->val;
    root->val = s;
    dfs(root->left);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func convertBST(root *TreeNode) *TreeNode {
  s := 0
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    dfs(root.Right)
    s += root.Val
    root.Val = s
    dfs(root.Left)
  }
  dfs(root)
  return root
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *   this.val = (val===undefined ? 0 : val)
 *   this.left = (left===undefined ? null : left)
 *   this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var convertBST = function (root) {
  let s = 0;
  function dfs(root) {
    if (!root) {
      return;
    }
    dfs(root.right);
    s += root.val;
    root.val = s;
    dfs(root.left);
  }
  dfs(root);
  return root;
};

Solution 2

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def convertBST(self, root: TreeNode) -> TreeNode:
    s = 0
    node = root
    while root:
      if root.right is None:
        s += root.val
        root.val = s
        root = root.left
      else:
        next = root.right
        while next.left and next.left != root:
          next = next.left
        if next.left is None:
          next.left = root
          root = root.right
        else:
          s += root.val
          root.val = s
          next.left = None
          root = root.left
    return node

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public TreeNode convertBST(TreeNode root) {
    int s = 0;
    TreeNode node = root;
    while (root != null) {
      if (root.right == null) {
        s += root.val;
        root.val = s;
        root = root.left;
      } else {
        TreeNode next = root.right;
        while (next.left != null && next.left != root) {
          next = next.left;
        }
        if (next.left == null) {
          next.left = root;
          root = root.right;
        } else {
          s += root.val;
          root.val = s;
          next.left = null;
          root = root.left;
        }
      }
    }
    return node;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  TreeNode* convertBST(TreeNode* root) {
    int s = 0;
    TreeNode* node = root;
    while (root) {
      if (root->right == nullptr) {
        s += root->val;
        root->val = s;
        root = root->left;
      } else {
        TreeNode* next = root->right;
        while (next->left && next->left != root) {
          next = next->left;
        }
        if (next->left == nullptr) {
          next->left = root;
          root = root->right;
        } else {
          s += root->val;
          root->val = s;
          next->left = nullptr;
          root = root->left;
        }
      }
    }
    return node;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func convertBST(root *TreeNode) *TreeNode {
  s := 0
  node := root
  for root != nil {
    if root.Right == nil {
      s += root.Val
      root.Val = s
      root = root.Left
    } else {
      next := root.Right
      for next.Left != nil && next.Left != root {
        next = next.Left
      }
      if next.Left == nil {
        next.Left = root
        root = root.Right
      } else {
        s += root.Val
        root.Val = s
        next.Left = nil
        root = root.Left
      }
    }
  }
  return node
}