DFS on Trees
Prereq: DFS Introduction
Think like a node
The key to solving tree problems using DFS is to think from the perspective of a node instead of looking at the whole tree. This is in line with how recursion is written. Reason from a node, decide how the current node should be proceeded, then recurse on children and let recursion takes care of the rest.
When you are a node, the only things you know are 1) your value and 2) how to get to your children. So the recursive function you write manipulates these things.
The template for DFS on tree is:
function dfs(node, state): if node is null: ... return left = dfs(node.left, state) right = dfs(node.right, state) ... return ...
Defining the recursive function
Two things we need to decide to define the function:
1. Return value (passing value up from child to parent)
What do we want to return after visiting a node. For example, for max depth problem this is max depth for the current node's subtree. If we are looking for a node in the tree, we'd want to return that node if found, else return null. Use return value to pass information from children to parent.
2. Identify state(s) (passing value down from parent to child)
What states do we need to maintain to compute the return value for the current node. For example, to know if the current node's value is larger than its parent we have to maintain the parent's value as a state. State becomes DFS's function arguments. Use states to pass information from parent to children.
Consider the problem of pretty-print a binary tree. Given directory tree
We want to "pretty-print" the directory structure with indents like this:
/ foo baz bar
We can pass the current indent level as a state of the recursive call.
indent_per_level = ' ' function dfs(node, indent_level): ... current_indent_level = indent_level + indent_per_level print(current_indent_level + node.val) dfs(node, current_indent_level)
Using return value vs. global variable
Consider the problem of finding the maximum value in a binary tree.
11 is the large value.
Using return value (divide and conquer)
One way to solve it is to use return value to pass the maximum value we have encountered back to parent node, and let the parent node compare it with the return value from the other child. This is more of a divide and conquer approach.
function dfs(node): if node is null: return MIN_VALUE left_max_val = dfs(node.left) right_max_val = dfs(node.right) return max(node.val, left_max_val, right_max_val)
Using global variable
Another way to solve it is to traverse the tree while keeping a global variable that keeps track of the maximum value we have encountered. After the dfs, we return the global variable.
... # global variable to record current max value # initialize to minimum value possible so any node will be larger max_val = MIN_VALUE function dfs(node): if node is null: return if node.val > max_val: # update the global variable if current value is larger max_val = root.val # recurse dfs(node.left) dfs(node.right) function get_max_val(root) dfs(root) # kick off dfs from root node return max_val
It's more of a personal preference which one you use. One could argue global variables are bad and therefore the divide and conquer. However, sometimes it's easier to use a global variable. Recall that divide and conquer has two steps - partition and merge. If the merge step is complex, then using a global variable might simplify things.