58 Balance a Binary Search Tree

題目

Given a binary search tree, return a balanced binary search tree with the same node values.

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

If there is more than one answer, return any of them.

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,null,null] is also correct.

Constraints:

The number of nodes in the tree is between 1 and 10^4.
The tree nodes will have distinct values between 1 and 10^5.

分析

題意：將二叉搜索樹平衡；每個節點的兩個子樹深度不超過一個節點，稱爲平衡。

一個笨辦法是先獲取原樹的所有值，然後根據二叉搜索樹的規則(二分法)，重新建立起一顆平衡的二叉搜索樹。

解答

class Solution {
    List<TreeNode> sortedArr = new ArrayList<>();
    public TreeNode balanceBST(TreeNode root) {
    	// 中序遍歷得到的就是排好序的
        inorderTraverse(root);
        // 通過二分法重新構造二叉搜索樹
        return sortedArrayToBST(0, sortedArr.size() - 1);
    }
    void inorderTraverse(TreeNode root) {
        if (root == null) return;
        inorderTraverse(root.left);
        sortedArr.add(root);
        inorderTraverse(root.right);
    }
    TreeNode sortedArrayToBST(int start, int end) {
        if (start > end) return null;
        int mid = (start + end) / 2;
        TreeNode root = sortedArr.get(mid);
        root.left = sortedArrayToBST(start, mid - 1);
        root.right = sortedArrayToBST(mid + 1, end);
        return root;
    }
}