# Posts published in “Tree”

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
1
/ \
2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
4
/ \
9   0
/ \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

## Solution: Recursion

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

## C++

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
/ \
9  20
/  \
15   7

## Solution: Recursion

Similar to LC 105

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

## Related Problems

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
/ \
9  20
/  \
15   7

## Solution: Recursion

Preprocessing: use a hashtable to store the index of element in preorder array.

For an element in inorder array, find the pos of it in preorder array in O(1), anything to the left will be the leftchild and anything to the right will be the right child.

e.g.
buildTree([9, 3, 15, 20, 7], [3, 9, 20, 15, 7]):
root = TreeNode(9) # inorder[0] = 9
root.left = buildTree([3], [3])
root.right = buildTree([15, 20, 7], [20, 15, 7])
return root

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

## Related Problems

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

1
/
3
\
2

Output: [3,1,null,null,2]

3
/
1
\
2


Example 2:

Input: [3,1,4,null,null,2]

3
/ \
1   4
/
2

Output: [2,1,4,null,null,3]

2
/ \
1   4
/
3


• A solution using O(n) space is pretty straight forward.
• Could you devise a constant space solution?

## Solution: Inorder traversal

Using inorder traversal to find two nodes that have val < prev.val

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

## C++

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1} Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.


Note:

• You may only use constant extra space.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

## Solution: Recursion

Do a preorder traversal:
1. return if self is empty or leaf
2. self.left->next = self.right
3. if self.next: self.right.next = self.next.left

Time complexity: O(n)
Space complexity: O(log(n)) since it’s a perfect tree

## C++

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