Largest BST Subtree
https://leetcode.com/problems/largest-bst-subtree/description/
Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.
Note:
A subtree must include all of its descendants.
Here's an example:
10 / \
5 15
/ \
1 8 7
The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree's size, which is 3.
Follow up:
Can you figure out ways to solve it with O(n) time complexity?
Thoughts
分治, 检查当前点是否满足BST条件, 即当左右子树都是BST时,node.val比左子树最大值大, 比右子树最小值小. 因此局部需要保存三个值, res[0]保存当前节点和子树构成bst时的size, 置为-1如果不能构成bst, res[1]保存能构成bst时整棵子树的最小值, res[2]保存子树最大值. 外加一个全局变量保存全局当前能构成的最大bst是多大.
Code
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int max = 0;
private int[] helper(TreeNode root) {
int[] res = new int[3]; // res, min, max
if (root == null) {
return res;
}
int[] left = helper(root.left);
int[] right = helper(root.right);
if (left[0] == -1 || right[0] == -1 || left[0] != 0 && root.val <= left[2] || right[0] != 0 && root.val >= right[1]) {
res[0] = -1;
return res;
}
res[0] = left[0] + right[0] + 1;
max = Math.max(max, res[0]);
res[1] = left[0] == 0 ? root.val : left[1];
res[2] = right[0] == 0 ? root.val : right[2];
return res;
}
public int largestBSTSubtree(TreeNode root) {
helper(root);
return max;
}
}
Analysis
时间复杂度O(N).
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