Convert Sorted Array to Binary Search Tree——LeetCode
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
题目大意:给定一个升序序列的数组,将其转换为二叉搜索树。
解题思路:数组中间元素是根元素,根元素将数组划分为两部分,两个部分的中心元素分别为根元素的左右孩子,依次递推。。。可用迭代或递归来做。
解法一(迭代):迭代需要自己额外记录下标
public TreeNode sortedArrayToBST(int[] num) {
if (num == null || num.length == 0) {
return null;
}
if (num.length == 1) {
return new TreeNode(num[0]);
}
Deque<Pair> queue = new ArrayDeque<>();
TreeNode root = new TreeNode(num[(num.length - 1) / 2]);
queue.add(new Pair(0, num.length - 1, root));
while (!queue.isEmpty()) {
Pair pair = queue.poll();
int pre = pair.pre;
int post = pair.post;
int curr = (post + pre) / 2;
TreeNode tNode = pair.node;
if (curr > pre) {
TreeNode left = new TreeNode(num[(curr - 1 + pre) / 2]);
tNode.left = left;
queue.add(new Pair(pre, curr - 1, left));
}
if (post > curr) {
TreeNode right = new TreeNode(num[(post + curr + 1) / 2]);
tNode.right = right;
queue.add(new Pair(curr + 1, post, right));
}
}
return root;
} private static class Pair {
int pre;
int post;
TreeNode node; public Pair(int pre, int post, TreeNode node) {
this.pre = pre;
this.post = post;
this.node = node;
}
}
解法二(递归):
public TreeNode sortedArrayToBST(int[] num) {
if (num == null || num.length == 0) {
return null;
}
TreeNode root = helper(0, num.length - 1, num);
return root;
} private TreeNode helper(int low, int high, int[] num) {
if (low > high) {
return null;
}
int mid = (low + high) / 2;
TreeNode node = new TreeNode(num[mid]);
node.left = helper(low, mid - 1, num);
node.right = helper(mid + 1, high, num);
return node;
}
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