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花花酱 LeetCode 1008. Construct Binary Search Tree from Preorder Traversal

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Note: 

  1. 1 <= preorder.length <= 100
  2. The values of preorder are distinct.

Solution: Recursion

Time complexity: O(n^2)
Space complexity: O(n)

C++

C++/iterator

花花酱 LeetCode 998. Maximum Binary Tree II

We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.

Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with the following Construct(A) routine:

  • If A is empty, return null.
  • Otherwise, let A[i] be the largest element of A.  Create a root node with value A[i].
  • The left child of root will be Construct([A[0], A[1], ..., A[i-1]])
  • The right child of root will be Construct([A[i+1], A[i+2], ..., A[A.length - 1]])
  • Return root.

Note that we were not given A directly, only a root node root = Construct(A).

Suppose B is a copy of A with the value val appended to it.  It is guaranteed that B has unique values.

Return Construct(B).

Example 1:

Input: root = [4,1,3,null,null,2], val = 5
Output: [5,4,null,1,3,null,null,2]
Explanation: A = [1,4,2,3], B = [1,4,2,3,5]

Example 2:

Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]
Explanation: A = [2,1,5,4], B = [2,1,5,4,3]

Example 3:

Input: root = [5,2,3,null,1], val = 4
Output: [5,2,4,null,1,3]
Explanation: A = [2,1,5,3], B = [2,1,5,3,4]

Note:

  1. 1 <= B.length <= 100

Solution: Recursion

Since val is the last element of the array, we compare root->val with val, if root->val > val then val can be inserted into the right subtree recursively, otherwise, root will be the left subtree of val.

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 993. Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Note:

  1. The number of nodes in the tree will be between 2 and 100.
  2. Each node has a unique integer value from 1 to 100.

Solution: Preorder traversal

Time complexity: O(n)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 988. Smallest String Starting From Leaf

Given the root of a binary tree, each node has a value from 0 to 25representing the letters 'a' to 'z': a value of 0 represents 'a', a value of 1 represents 'b', and so on.

Find the lexicographically smallest string that starts at a leaf of this tree and ends at the root.

(As a reminder, any shorter prefix of a string is lexicographically smaller: for example, "ab" is lexicographically smaller than "aba".  A leaf of a node is a node that has no children.)

Example 1:

Input: [0,1,2,3,4,3,4]
Output: "dba"

Example 2:

Input: [25,1,3,1,3,0,2]
Output: "adz"

Example 3:

Input: [2,2,1,null,1,0,null,0]
Output: "abc"

Note:

  1. The number of nodes in the given tree will be between 1 and 1000.
  2. Each node in the tree will have a value between 0 and 25.

Solution: Recursion

Time complexity: O(n^2)
Space complexity: O(n^2)

C++

Python3

花花酱 LeetCode 987. Vertical Order Traversal of a Binary Tree

Given a binary tree, return the vertical order traversal of its nodes values.

For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).

Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).

If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.

Return an list of non-empty reports in order of X coordinate.  Every report will have a list of values of nodes.

Example 1:

Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation: 
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).

Example 2:

Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation: 
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6.

Note:

  1. The tree will have between 1 and 1000 nodes.
  2. Each node’s value will be between 0 and 1000.

Solution: Ordered Map+ Ordered Set

Time complexity: O(nlogn)
Space complexity: O(n)

C++

Python3