0654. Maximum Binary Tree
Last updated
Last updated
**Input:** nums = [3,2,1,6,0,5]
**Output:** [6,3,5,null,2,0,null,null,1]
**Explanation:** The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
- The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
- Empty array, so no child.
- The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
- Empty array, so no child.
- Only one element, so child is a node with value 1.
- The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
- Only one element, so child is a node with value 0.
- Empty array, so no child.**Input:** nums = [3,2,1]
**Output:** [3,null,2,null,1]/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode constructMaximumBinaryTree(int[] nums) {
return build(nums, 0, nums.length-1);
}
public TreeNode build(int[] nums, int start, int end) {
if (start > end) return null;
// find max
int maxIdx = -1, max = Integer.MIN_VALUE;
for (int i = start; i <= end; i++) {
if (nums[i] > max) {
max = nums[i];
maxIdx = i;
}
}
// build
TreeNode root = new TreeNode(max);
root.left = build(nums, start, maxIdx-1);
root.right = build(nums, maxIdx+1, end);
return root;
}
}
/*
just recursion build the subtree
*/