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 - Queue

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 - Recursion

public class Solution {
    int max=0;
 
    public int longestConsecutive(TreeNode root) {
        helper(root);
        return max;
    }
 
    public int helper(TreeNode root) {
        if(root==null)
            return 0;
 
        int l = helper(root.left);
        int r = helper(root.right);
 
        int fromLeft = 0;
        int fromRight= 0;
 
        if(root.left==null){
            fromLeft=1;
        }else if(root.left.val-1==root.val){
            fromLeft = l+1;
        }else{
            fromLeft=1;
        }
 
        if(root.right==null){
            fromRight=1;
        }else if(root.right.val-1==root.val){
            fromRight = r+1;
        }else{
            fromRight=1;
        }
 
        max = Math.max(max, fromLeft);
        max = Math.max(max, fromRight);
 
        return Math.max(fromLeft, fromRight);
    }
 
}
Category >> Algorithms  
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  1. Avinash Sharma on 2016-9-17

    I believe there is a mistake in first solution in the part of the code where:

    if(head.val = head.left.val-1) it should be if(head.val == head.left.val+1) as the left value should be one less only then consecutive sequence makes sense.

  2. Aditya Vutukuri on 2016-10-27

    I guess there is a mistake. The tree here is considered as BST instead of BT. The sequence could be increasing or decreasing.

  3. Subodh Karwa on 2016-11-10

    No use of calculating the max and needs to be ignored

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

  4. Venk on 2016-12-17

    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);
    }

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