Last Updated: 15 Oct, 2020
Deepest Leaves Sum
Easy
Asked in companies
You have been given a binary tree of integers. Your task is to calculate the sum of all the leaf nodes which are present at the deepest level of this binary tree. If there are no such nodes, print 0.
NOTE: The deepest level of a binary tree is the level which is present at the maximum depth from the root node.
Follow up:
Can you solve this problem in a single traversal of the binary tree?
Input Format:
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.
The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 on its place.
For example, the input for the tree depicted in the below image will be:
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).
Note :
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Output Format:
For each test case, print the sum of all the leaf nodes present at the deepest level of the given binary tree.
Print the output of each test case in a separate line.
Note:
You are not required to print the expected output, it has already been taken care of. Just implement the function.
Constraints:
1 <= T <= 10^2
0 <= N <= 3000
Where N is the number of nodes in the binary tree
The sum will always fit in a 32-bit integer.
Time Limit: 1 sec
Approaches
The idea is pretty simple. First, we will find the deepest level of the given binary tree, say ‘maxDepth’. We can do this with a single traversal of the binary tree.
After finding the deepest level of the binary tree, we will again traverse the given tree. Also, we will maintain a ‘sum’ and ‘currLevel’ variable for each node in the tree. Whenever the ‘currLevel’ which represents the level of the current node, is equal to the ‘maxDepth’, we will add this node in our ‘sum’ variable.
At last, the value of ‘sum’ will be our sum of all the nodes which are present in the deepest level of the binary tree.
In approach 1, we are finding the deepest level of the binary tree and then exploring it in a separate traversal. We do not need two separate traversals for both these parts.
While traversing for finding the deepest level in the binary tree, we can maintain a variable ‘sum’ to add the nodes which are at the same level as the deepest level.
While traversing-
- If we find a node which is present at a deeper level, we will reset the value of the ‘sum’ and set ‘maxDepth’ to ‘currLevel’.
- If a node is at the same level as the ‘maxDepth’, we will simply add this node to the ‘sum’.
The idea is to perform a level order traversal of the given binary tree from the root node. The nodes which are present at the deepest level will be processed last as we are processing the tree level by level.
Here is the complete algorithm-
- Declare ‘ans’ variable and initialize to 0.
- Declare a queue and push the root node into the queue.
- Loop till the queue is not empty.
- Initialise the ‘ans’ to 0, as a deeper level is being processed.
- Process all the nodes of the queue, which represent the current level of the binary tree and add them to the ‘ans’.
- While processing the nodes, push their children, if any, into the queue.
- Return ‘ans’.
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