Leetcode: Verify Preorder Sequence in Binary Search Tree

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree.

You may assume each number in the sequence is unique.

Follow up:
Could you do it using only constant space complexity?

           4

        /     \

      2         6

    /   \      /   \

  1     3   5      7

如图所示BST， preorder结果是 4 2 1 3 6 5 7

4是root, 213是左子树，657是右子树

idea就是右子树不能再出现比root小的数，一旦出现，return false. 这样一层一层recursion, 直到子树只剩一个node， return true

 1 public class Solution {
 2     public boolean verifyPreorder(int[] preorder) {
 3         if(preorder==null || preorder.length==0) return true;
 4         return helper(0, preorder.length-1, preorder);
 5     }
 6     
 7     public boolean helper(int l, int r, int[] preorder) {
 8         if (l >= r) return true;
 9         int root = l;
10         int right = l-1;
11         boolean found = false;
12         for (int i=l+1; i<=r; i++) {
13             if (preorder[i] > preorder[root]) { //found right subtree 1st node
14                 if (!found) {
15                     found = true;
16                     right = i;
17                 }
18             }
19             else {
20                 if (found) return false;
21             }
22         }
23         if (!found) return helper(root+1, r, preorder);
24         else return helper(root+1, right-1, preorder) && helper(right, r, preorder);
25     }
26 }

这样时间代价很大

更好的方法，参考https://leetcode.com/discuss/51543/java-o-n-and-o-1-extra-space

Time: O(N), Space: O(H)

imulate the traversal, keeping a stack of nodes. If the next number is smaller than the last stack value, then we're still in the left subtree of all stack nodes, so just push the new one onto the stack. Otherwise, pop all smaller ancestor values, as we must now be in their right subtrees (or even further, in the right subtree of an ancestor). Also, use the popped values as a lower bound, since being in their right subtree means we must never come across a smaller number anymore.

 1 public boolean verifyPreorder(int[] preorder) {
 2     int low = Integer.MIN_VALUE;
 3     Stack<Integer> path = new Stack();
 4     for (int p : preorder) {
 5         if (p < low)
 6             return false;
 7         while (!path.empty() && p > path.peek())
 8             low = path.pop();
 9         path.push(p);
10     }
11     return true;
12 }

Solution 2 ... O(1) extra space

Same as above, but use the given array as the stack. i refer to the stack.top.next

 1 public boolean verifyPreorder(int[] preorder) {
 2     int low = Integer.MIN_VALUE, i = 0;
 3     for (int p : preorder) {
 4         if (p < low)
 5             return false;
 6         while (i > 0 && p > preorder[i-1])
 7             low = preorder[--i];
 8         preorder[i++] = p;
 9     }
10     return true;
11 }

 i refer to stack.top

 1 public boolean verifyPreorder(int[] preorder) {
 2     int low = Integer.MIN_VALUE, i = -1;
 3     for (int p : preorder) {
 4         if (p < low)
 5             return false;
 6         while (i >= 0 && p > preorder[i])
 7             low = preorder[i--];
 8         preorder[++i] = p;
 9     }
10     return true;
11 }

 

转载于:https://www.cnblogs.com/EdwardLiu/p/5066035.html