Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
Given a binary tree
1 / \ 2 3 / \ 4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
Any path can be written as two arrows (in different directions) from some node, where an arrow is a path that starts at some node and only travels down to child nodes.\n
If we knew the maximum length arrows
L, R for each child, then the best path touches
L + R + 1 nodes.
Let\'s calculate the depth of a node in the usual way: max(depth of node.left, depth of node.right) + 1. While we do, a path "through" this node uses 1 + (depth of node.left) + (depth of node.right) nodes. Let\'s search each node and remember the highest number of nodes used in some path. The desired length is 1 minus this number.\n\n
Time Complexity: . We visit every node once.\n
Space Complexity: , the size of our implicit call stack during our depth-first search.\n
Analysis written by: @awice.\n