# LeetCode – Binary Tree Longest Consecutive Sequence (Java)

Given a binary tree, find the length of the longest consecutive sequence path.

The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from parent to child (cannot be the reverse).

Java Solution 1 - BFS

```public int longestConsecutive(TreeNode root) { if(root==null) return 0;   LinkedList<TreeNode> nodeQueue = new LinkedList<TreeNode>(); LinkedList<Integer> sizeQueue = new LinkedList<Integer>();   nodeQueue.offer(root); sizeQueue.offer(1); int max=1;   while(!nodeQueue.isEmpty()){ TreeNode head = nodeQueue.poll(); int size = sizeQueue.poll();   if(head.left!=null){ int leftSize=size; if(head.val==head.left.val-1){ leftSize++; max = Math.max(max, leftSize); }else{ leftSize=1; }   nodeQueue.offer(head.left); sizeQueue.offer(leftSize); }   if(head.right!=null){ int rightSize=size; if(head.val==head.right.val-1){ rightSize++; max = Math.max(max, rightSize); }else{ rightSize=1; }   nodeQueue.offer(head.right); sizeQueue.offer(rightSize); }     }   return max; }```

Java Solution 2 - DFS

```class Solution { int max;   public int longestConsecutive(TreeNode root) { helper(root); return max; }   private int helper(TreeNode t){ if(t==null){ return 0; }   int leftMax = helper(t.left); int rightMax = helper(t.right);   int leftTotal = 0; if(t.left == null){ leftTotal = 1; }else if(t.val+1 == t.left.val){ leftTotal = leftMax+1; }else{ leftTotal = 1; }   int rightTotal = 0; if(t.right == null){ rightTotal = 1; }else if(t.val+1 == t.right.val){ rightTotal = rightMax+1; }else{ rightTotal = 1; }   max = Math.max(max, leftTotal); max = Math.max(max, rightTotal);   int longer = Math.max(leftTotal, rightTotal);   return longer; } }```
Category >> Algorithms
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• Do you assume that the values of the tree are unique?

• p.andrey

``` public int longestConsecutive(Node root) { if (root == null) return 1;```

``` return preOrder(root, null, -1); } private int preOrder(Node n, Node p, int pCurr) { int cMax = 1; if(p != null && n.value == p.value + 1) cMax = pCurr + 1; int max = cMax; if (n.left != null) { int childMax = preOrder(n.left, n, cMax); max = Math.max(max, childMax); } if (n.right != null) { int childMax = preOrder(n.left, n, cMax); max = Math.max(max, childMax); } ```

``` return max; } ```

• Zia Khalid

yes, you are correct there is mistake in first solution.

• Venk

Simple solution:

``` static int max = 1; private static int getMax(TreeNode root) {```

``` helper(root, 1); return max; } private static void helper(TreeNode root, int count) { if(root == null) return; max = Math.max(count, max); if(root.left != null) helper(root.left, (root.val + 1 == root.left.val) ? count + 1 : 1); ```

``` if(root.right != null) helper(root.right, (root.val + 1 == root.right.val) ? count + 1 : 1); } ```

• Subodh Karwa

No use of calculating the max and needs to be ignored

max = Math.max(max, fromLeft);
max = Math.max(max, fromRight);