0654. Maximum Binary Tree
Medium | Tree + Monotonic Stack | 168 ms (99.67%), 14.9 MB (64.90%)
Source: LeetCode - Maximum Binary Tree GitHub: Solution / Performance
You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:
Create a root node whose value is the maximum value in
nums.Recursively build the left subtree on the subarray prefix to the left of the maximum value.
Recursively build the right subtree on the subarray suffix to the right of the maximum value.
Return the maximum binary tree built from nums.
Constraints:
1 <= nums.length <= 10000 <= nums[i] <= 1000All integers in
numsare unique.
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.Recursive Solution
We need two functions (1) Helper for constructing a binary tree (2) Find the MAX element using the left and right boundaries.
The helper function recursively calls itself and the termination condition is when left-boundary >= right-boundary.
Iterative Solution Iterate through the input integer array and use stack to record previous nodes
current node's value > previous node's value Keep popping from the stack and append the last node to the current node's left
current node's value < previous node's value Append the current node to the previous node's right
Push current node onto the stack
Last updated
Was this helpful?