654. Maximum Binary Tree
Tags: ‘Tree’
Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:
- The root is the maximum number in the array.
- The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
- The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.
Construct the maximum tree by the given array and output the root node of this tree.
Example 1:
Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:
6
/ \
3 5
\ /
2 0
\
1
Note:
- The size of the given array will be in the range [1,1000].
Solution
// Recursive
// O(n^2) O(n) 99% 97%
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
if (nums == null) return null;
return build(nums, 0, nums.length - 1);
}
private TreeNode build(int[] nums, int start, int end) {
if (start > end) return null; // End Case
int maxIndex = findMaxIndex(nums, start, end);
TreeNode node = new TreeNode(nums[maxIndex]);
node.left = build(nums, start, maxIndex -1);
node.right = build(nums, maxIndex + 1, end);
return node;
}
private int findMaxIndex(int[] nums, int start, int end) {
int max = start;
for (int i = start+ 1; i <= end; i++) {
if (nums[i] > nums[max]) max = i;
}
return max;
}
}
// O(n) solution 12% 78%
// Heap is slower than stack
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
Deque<TreeNode> stack = new LinkedList<>();
for(int i = 0; i < nums.length; i++) {
TreeNode curr = new TreeNode(nums[i]);
while(!stack.isEmpty() && stack.peek().val < nums[i]) {
curr.left = stack.pop();
}
if(!stack.isEmpty()) {
stack.peek().right = curr;
}
stack.push(curr);
}
return stack.isEmpty() ? null : stack.removeLast();
}
}