Given a Binary Tree, find the vertical sum of the nodes that are in the same vertical line. Print all sums through different vertical lines.
1 / \ 2 3 / \ / \ 4 5 6 7
The tree has 5 vertical lines
Vertical-Line-1 has only one node 4 => vertical sum is 4
Vertical-Line-2: has only one node 2=> vertical sum is 2
Vertical-Line-3: has three nodes: 1,5,6 => vertical sum is 1+5+6 = 12
Vertical-Line-4: has only one node 3 => vertical sum is 3
Vertical-Line-5: has only one node 7 => vertical sum is 7
So expected output is 4, 2, 12, 3 and 7
We need to check the Horizontal Distances from the root for all nodes. If two nodes have the same Horizontal Distance (HD), then they are on the same vertical line. The idea of HD is simple. HD for root is 0, a right edge (edge connecting to right subtree) is considered as +1 horizontal distance and a left edge is considered as -1 horizontal distance. For example, in the above tree, HD for Node 4 is at -2, HD for Node 2 is -1, HD for 5 and 6 is 0 and HD for node 7 is +2.
We can do an in-order traversal of the given Binary Tree. While traversing the tree, we can recursively calculate HDs. We initially pass the horizontal distance as 0 for root. For left subtree, we pass the Horizontal Distance as Horizontal distance of root minus 1. For right subtree, we pass the Horizontal Distance as Horizontal Distance of root plus 1.
Following is Java implementation for the same. HashMap is used to store the vertical sums for different horizontal distances. Thanks to Nages for suggesting this method.
Time Complexity: O(n)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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- Print a Binary Tree in Vertical Order | Set 3 (Using Level Order Traversal)
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- Convert a Binary Tree to Threaded binary tree | Set 2 (Efficient)