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LeetCode-654-Maximum-Binary-Tree.java
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LeetCode-654-Maximum-Binary-Tree.java
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/*
https://github.com/drinkbeer/CodingQuestion/blob/master/cartesian-tree.md
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
// 1. Recursively. Build Cartesian Tree
/*
Avg Time O(NlogN)
Worst Time O(N^2), when the array is ascending and descending
*/
// public TreeNode constructMaximumBinaryTree(int[] nums) {
// if (nums == null || nums.length == 0) return null;
// return build(nums, 0, nums.length - 1);
// }
// private TreeNode build(int[] nums, int s, int e) {
// if (s > e) return null;
// int max = -1, maxVal = Integer.MIN_VALUE;
// for (int i = s; i <= e; i++) {
// if (nums[i] > maxVal) {
// maxVal = nums[i];
// max = i;
// }
// }
// TreeNode node = new TreeNode(nums[max]);
// node.left = build(nums, s, max - 1);
// node.right = build(nums, max + 1, e);
// return node;
// }
// 2. Using a Deque based stack
/*
https://leetcode.com/problems/maximum-binary-tree/discuss/106156/Java-worst-case-O(N)-solution
Runtime: 8 ms, faster than 14.24% of Java online submissions for Maximum Binary Tree.
Memory Usage: 39.4 MB, less than 79.45% of Java online submissions for Maximum Binary Tree.
Worst Time O(N)
*/
public TreeNode constructMaximumBinaryTree(int[] nums) {
if (nums == null || nums.length == 0) return null;
Deque<TreeNode> stack = new ArrayDeque<>();
for (int i = 0; i < nums.length; i++) {
TreeNode curr = new TreeNode(nums[i]);
// traverse the nodes in the stack, and find the max-smaller one than nums[i], make it to be the left node of curr
while(!stack.isEmpty() && stack.peek().val < nums[i]) {
curr.left = stack.pop();
}
if (!stack.isEmpty() && stack.peek().val > nums[i]) {
stack.peek().right = curr;
}
stack.push(curr);
}
return stack.isEmpty() ? null : stack.peekLast();
}
}