669. Trim a Binary Search Tree

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.

Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.

题目解析：

• 递归裁剪

代码：

class Solution {
    public TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) return root;
        if (root.val < low) {
            return trimBST(root.right, low, high);
        } else if (root.val > high) {
            return trimBST(root.left, low, high);
        } else {
            root.left = trimBST(root.left, low, high);
            root.right = trimBST(root.right, low, high);
            return root;
        }
    }
}

108. Convert Sorted Array to Binary Search Tree

Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.

题目解析：

• 递归构造

代码：

lass Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        return build(nums, 0, nums.length - 1);
    }

    public TreeNode build(int[] nums, int start, int end) {
        if (start > end) return null;
        int mid = (start + end) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        root.left = build(nums, start, mid - 1);
        root.right = build(nums, mid + 1, end);
        return root;
    }
}

538. Convert BST to Greater Tree

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.

题目解析：

• 按照右中左的顺序遍历即可

代码：

class Solution {
    int sum = 0;
    public TreeNode convertBST(TreeNode root) {
        if (root == null) return null;
        convertBST(root.right);
        sum += root.val;
        root.val = sum;
        convertBST(root.left);
        return root;
    }
}

总结

1. 二叉树问题一般迭代法比递归简单
2. 根据不同的问题选择DFS（前中后序遍历）或者BFS
3. BST是从左到右排序，所以双指针也可以用来解决某些问题