We define a "root$-$to$-$node path" to be any sequence of nodes in a tree starting with the root and proceeding downward to a given node. The "root$-$to$-$node path sum" for that path is the sum of the values for all the nodes $($including the root and the given node$)$ along that path. Define an empty tree to contain no root$-$to$-$node paths $($and so its sum is zero$).$ Define a tree with one node $($equivalently$,$ the path to the root$)$ to have a root$-$to$-$node path consisting of just the root $($and so its sum is the value of the root$).$ Given a binary tree and a value sum, return true if the tree has some root$-$to$-$node path such that adding up all the values along the path equals sum. Return false if no such path exists.

Here are methods that you can use on the BinNode objects:

interface BinNode $\{$

public int value$()$;

public void setValue$($int v$)$;

public BinNode left$()$;

public BinNode right$()$;

public boolean isLeaf$()$;

$\}$

Your Answer:

public boolean hasPathSum$($BinNode root, int sum$)\{$

if $($root $==$ null$)\{$

return false;

$\}$

if $($root$.$isLeaf$())\{$

return root.value$()==$ sum;

$\}$

int remainingSum $=$ sum $-$ root.value$()$;

return hasPathSum$($root$.$left$(),$ remainingSum$)||$ hasPathSum$($root$.$right$(),$ remainingSum$)$;

$\}$

public boolean BTpathsum$($BinNode root, int sum$)$

$\{$

$\}$