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1382. Balance a Binary Search Tree

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.
 * 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.
# 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.
 * 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);
}

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