思路1:暴力法,遍历当前待检查数组,找到第一个大于数组起始位置的位置i,则 i 为右子树根节点,然后递归判断左右子树。对于根节点,最坏情况下要遍历整个数组,时间为O(N),因为要递归检查每个节点,因此总的时间复杂度是O(N^2)。
思路2:参考StefanPochmann大神的作品 https://leetcode.com/discuss/51543/java-o-n-and-o-1-extra-space, 时间复杂度是O(N),空间复杂度是O(h)。
public class Solution { // Method 2 public boolean verifyPreorder(int[] preorder) { if (preorder == null || preorder.length == 0) return true; LinkedList<Integer> stack = new LinkedList<Integer>(); int low = Integer.MIN_VALUE; for (int i = 0; i < preorder.length; i++) { if (preorder[i] < low) return false; while (!stack.isEmpty() && preorder[i] > stack.peek()) { low = stack.pop(); } stack.push(preorder[i]); } return true; } // Method 1: brute force public boolean verifyPreorder1(int[] preorder) { if (preorder == null) return false; return recur(preorder, 0, preorder.length - 1); } public boolean recur(int[] preorder, int start, int end) { if (start >= end) return true; int i = start + 1; while (i <= end && preorder[i] < preorder[start]) i++; if (!recur(preorder, start + 1, i - 1)) return false; int rightStart = i; while (i <= end && preorder[i] > preorder[start]) i++; if (i <= end) return false; return recur(preorder, rightStart, end); } }