## Tuesday, November 29, 2016

### Binary Tree All Paths, Max Sum Path, Diameter, Longest Path, Shortest Path, Closest Leaf

http://www.zrzahid.com/binary-tree-all-paths-max-sum-path-diameter-longest-path-shortest-path-closest-leaf/
Given a Binary Tree T.
1. Find all paths from root to each of the leaves in T.
2. Find the longest path from root to a leaf (also called Max Depth or Height of the tree).
3. Find the path that has largest sum along the path in T.
4. Find distance (shortest) between given two nodes in T.
5. Diameter of T or Longest path between any two nodes in T.
6. Find mirror image tree of T.
7. Find min length path that sums to a given value.
8. Find the closest leaf from root.
For example, in the following
```                1
/   \
2     3
/    /   \
4    5     6
\        /
7      8
```
1. All Paths – [1–>2–>4–>7, 1–>3–>5, 1–>3–>6–>8]
2. Longest path aka Max Depth aka height of the tree is 4 because of longest path [1–>2–>4–>7] or [1–>3–>6–>8]
3. Max sum path – 31 from path [7–>4–>2->1–>3–>6–>8]
4. Distance between Node 5 and Node 8 is 3.
5. Diameter of T is 7 because there are 7 nodes in the longest path between node 7 and node 8.
6. Mirror image tree of T is
```                1
/   \
3     2
/ \     \
6   5     4
\        /
8      7
```
7. Checkout previous post on Min len sum-path.
8. Closest leaf is 5 (1->3->5)
All Paths from Root to Leaves
We can do DFS traverse from root until we find a leaf node and print the path DFS traverse visited.
```public static void paths(BTNode root, String path, ArrayList<String> paths){
if(root == null){
return;
}

path=path+(path.isEmpty()? "" : "-->")+root.key;

if(root.left == null && root.right == null){
System.out.println("path > "+path);
return;
}

paths(root.left, path, paths);
paths(root.right, path, paths);
}
```

Largest Sum Path
We can do a similar DFS traverse. let’s take a binary tree.
Path with max sum can be formed by either –
```1. Starting from one subtree of root and going through root to other subtree (path 1), or
2. Path that doesn't go through root i.e. it ends at root. We can have 2 such paths -
2.1. Paths completely in left subtree that ends at root (path 2)
2.2. Paths completely in right subtree that ends at root (path 3)
```
So, at a node maximum path would be maximum of 3 paths (path1, path2, and path3). We can compute path2 and path3 by recursively in the respective subtrees. But how do we calculate path1 that spans across the left and right subtrees? Idea is that, as recursive calls to left and right sum will always return maximum of path2 or path3, so we can compute path1 using path2 and path3 recursively by updating a global max with the path1 computed.
```path2 or path3 computed as follows -
local_max(node) = max{local_max(node.left), local_max(node.right)} + node.key

Then , global maxima path at current node would be -

global_max = max{global_max, local_max(root.left)+local_max(root.left)+root.key}
```
```public static int maxSumPath1(BTNode node, int[] maxSum){
if(node ==  null){
return 0;
}

int leftSum = maxSumPath(node.left, maxSum);
int rightSum = maxSumPath(node.right, maxSum);

//update global max so far
//we take max by either going through current root or not going through
//current root if we take root then the max sum path goes from a left
//subtree node through root to a right subtree node
//if we don't take current root then we disregard path
//through root as if we have max path in either left or right subtree
maxSum[0] = Math.max(maxSum[0], leftSum+rightSum+node.key);

//return max in this current path that starts or ends at current root and doesn't go through it
return Math.max(leftSum, rightSum) + node.key;
}
```

Max depth or Height of Binary Tree
This is same as the number of nodes in the longest path. We can see that longest path with just one node is 1 i.e. the node itself. So, height aka max depth aka longest path is one more than the max of left or right sub tree height.
```height(T) = max{height(T.left), height(T.right)}+1;
```
```public static int maxDepth(BTNode root) {

if(root == null){
return 0;
}

int leftDepth = maxDepth(root.left);
int rightDepth = maxDepth(root.right);

return Math.max(leftDepth, rightDepth)+1;
}
```

Find distance between given two nodes
Distance between two nodes is the minimum number of edges to be traversed to reach one node from other. Note that, any two node in the tree must have a common ancestor. So, we can compute Lowest Common Ancestor (LCA) for the given two nodes and then we can find distance between two nodes as follows –
```lca = LCA(root, a, b);
d(a, b) = d(root, a)+d(root, b)-2*d(root, lca) , which is equivalent to
d(a,b) = level(a)+level(b)-2*level(lca).
```
So, basically we need two helper function to compute LCA and Level of the nodes. Below is an O(n) time algorithm
``` public static int getLevel(BTNode root, int count, BTNode node){
if(root == null){
return 0;
}

if(root == node){
return count;
}

int leftLevel = getLevel(root.left, count+1, node);
if(leftLevel != 0){
return leftLevel;
}
int rightLevel = getLevel(root.right, count+1, node);
return rightLevel;
}

public static BTNode LCA(BTNode root, BTNode x, BTNode y) {
if (root == null) return null;
if (root == x || root == y) return root;

BTNode leftSubTree = LCA(root.left, x, y);
BTNode rightSubTree = LCA(root.right, x, y);

//x is in one subtree and and y is on other subtree of root
if (leftSubTree != null && rightSubTree != null) return root;
//either x or y is present in one of the subtrees of root or none present in either side of the root
return leftSubTree!=null ? leftSubTree : rightSubTree;
}

public static int shortestDistance(BTNode root, BTNode a, BTNode b){
if(root == null){
return 0;
}

BTNode lca = LCA(root, a, b);
//d(a,b) = d(root,a) + d(root, b) - 2*d(root, lca)
return getLevel(root, 1, a)+getLevel(root, 1, b)-2*getLevel(root, 1, lca);
}
```

Diameter of Binary Tree
The diameter of a tree is the number of nodes on the longest path between any two leaves in the tree.
The diameter of a tree T is the largest of the following quantities:
* the diameter of T’s left subtree
* the diameter of T’s right subtree
* the longest path between leaves that goes through the root of T (this can be computed from the heights of the subtrees of T)
```D(T) = max{ D(T.left), D(T.right), LongestPath(T.root)}
LongestPath(T.root) = 1+height(T.root..left)+height(T.root.right)
height(T) = 1+max{height(T.left) , height(T.right) }
```
```public static int diameter(BTNode root){
//D(T) = max{D(T.left), D(T.right), LongestPathThrough(T.root)}
if(root == null){
return 0;
}

int leftHeight = maxDepth(root.left);
int rightHeight = maxDepth(root.right);

int leftDiameter = diameter(root.left);
int rightDiameter = diameter(root.right);

return Math.max(Math.max(leftDiameter, rightDiameter), leftHeight+rightHeight+1);
}
```

However, we can save some height computation overhead by computing the height while traversing the tree down from root. Below is a O(n) implementation of diameter.
```public static int diameter(BTNode root, int[] height){
if(root == null){
height[0] = 0;
return 0;
}

int[] leftHeight = {0}, rightHeight = {0};
int leftDiam = diameter(root.left, leftHeight);
int rightDiam = diameter(root.right, rightHeight);

height[0] = Math.max(leftHeight[0],rightHeight[0])+1;

return Math.max(Math.max(leftDiam, rightDiam), leftHeight[0]+rightHeight[0]+1);
}
```

Mirror Tree
swap left and right subtree recursively in tail recursion manner.
```public static void mirrorTree(BTNode root){
if(root == null){
return;
}

mirrorTree(root.left);
mirrorTree(root.right);

BTNode temp = root.right;
root.right = root.left;
root.left = temp;
}
```

Closest Leaf
Idea is pretty straightforward traverse left sub tree in DFS to find closest leaf. Then we backtrack to to root and traverse ins right subtree for closest leaf. Once both subtree are done we return minimum path length. Whenever we reach a leaf node we update a global max based on returned closest path length of left and right subtree. We also keep track of the closest leaf.
```public static BTNode closestLeaf(BTNode root, int[] len, BTNode closest){
if(root == null){
return closest;
}

len[0]++;
closest = closestLeaf(root.left, len, closest);
if(root.left == null && root.right == null){
if(len[0] < len[1]){
len[1] = len[0];
closest = root;
}
}
len[0]--;
closest = closestLeaf(root.right, len, closest);

return closest;
}
```