Question

1382. Balance a Binary Search Tree

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Description

Given the root of a binary search tree, return _a balanced binary search tree with the same node values_. If there is more than one answer, return any of them.

A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than 1.

 

Example 1:

Input: root = [1,null,2,null,3,null,4,null,null]
Output: [2,1,3,null,null,null,4]
Explanation: This is not the only correct answer, [3,1,4,null,2] is also correct.

Example 2:

Input: root = [2,1,3]
Output: [2,1,3]

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• 1 <= Node.val <= 105

Solutions

Solution 1: In-order Traversal + Construct Balanced Binary Search Tree

Since the original tree is a binary search tree, we can save the result of the in-order traversal in an array $nums$. Then we design a function $build(i, j)$, which is used to construct a balanced binary search tree within the index range $[i, j]$ in $nums$. The answer is $build(0, |nums| - 1)$.

The execution logic of the function $build(i, j)$ is as follows:

• If $i > j$, then the balanced binary search tree is empty, return an empty node;
• Otherwise, we take $mid = (i + j) / 2$ as the root node, then recursively build the left and right subtrees, and return the root node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the binary search tree.

# 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 balanceBST(self, root: TreeNode) -> TreeNode:
    def dfs(root: TreeNode):
      if root is None:
        return
      dfs(root.left)
      nums.append(root.val)
      dfs(root.right)

    def build(i: int, j: int) -> TreeNode:
      if i > j:
        return None
      mid = (i + j) >> 1
      left = build(i, mid - 1)
      right = build(mid + 1, j)
      return TreeNode(nums[mid], left, right)

    nums = []
    dfs(root)
    return build(0, len(nums) - 1)

/**
 * 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 List<Integer> nums = new ArrayList<>();

  public TreeNode balanceBST(TreeNode root) {
    dfs(root);
    return build(0, nums.size() - 1);
  }

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

  private TreeNode build(int i, int j) {
    if (i > j) {
      return null;
    }
    int mid = (i + j) >> 1;
    TreeNode left = build(i, mid - 1);
    TreeNode right = build(mid + 1, j);
    return new TreeNode(nums.get(mid), left, right);
  }
}

/**
 * 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* balanceBST(TreeNode* root) {
    dfs(root);
    return build(0, nums.size() - 1);
  }

private:
  vector<int> nums;

  void dfs(TreeNode* root) {
    if (!root) {
      return;
    }
    dfs(root->left);
    nums.push_back(root->val);
    dfs(root->right);
  }

  TreeNode* build(int i, int j) {
    if (i > j) {
      return nullptr;
    }
    int mid = (i + j) >> 1;
    TreeNode* left = build(i, mid - 1);
    TreeNode* right = build(mid + 1, j);
    return new TreeNode(nums[mid], left, right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func balanceBST(root *TreeNode) *TreeNode {
  ans := []int{}
  var dfs func(*TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    dfs(root.Left)
    ans = append(ans, root.Val)
    dfs(root.Right)
  }
  var build func(i, j int) *TreeNode
  build = func(i, j int) *TreeNode {
    if i > j {
      return nil
    }
    mid := (i + j) >> 1
    left := build(i, mid-1)
    right := build(mid+1, j)
    return &TreeNode{Val: ans[mid], Left: left, Right: right}
  }
  dfs(root)
  return build(0, len(ans)-1)
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function balanceBST(root: TreeNode | null): TreeNode | null {
  const nums: number[] = [];
  const dfs = (root: TreeNode | null): void => {
    if (root == null) {
      return;
    }
    dfs(root.left);
    nums.push(root.val);
    dfs(root.right);
  };
  const build = (i: number, j: number): TreeNode | null => {
    if (i > j) {
      return null;
    }
    const mid: number = (i + j) >> 1;
    const left: TreeNode | null = build(i, mid - 1);
    const right: TreeNode | null = build(mid + 1, j);
    return new TreeNode(nums[mid], left, right);
  };
  dfs(root);
  return build(0, nums.length - 1);
}