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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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