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articles/binary-tree-longest-consecutive-sequence-ii.md

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+## 1. Brute Force (Time Limit Exceeded)
+
+### Time & Space Complexity
+
+- Time complexity: $O(n^3)$
+- Space complexity: $O(n^3)$
+
+>  Where $n$ is the number of nodes in the input tree 
+
+---
+
+## 2. Single traversal
+
+::tabs-start
+
+```python
+class Solution:
+    def longestConsecutive(self, root: TreeNode) -> int:
+
+        def longest_path(root: TreeNode) -> List[int]:
+            nonlocal maxval
+
+            if not root:
+                return [0, 0]
+
+            inr = dcr = 1
+            if root.left:
+                left = longest_path(root.left)
+                if (root.val == root.left.val + 1):
+                    dcr = left[1] + 1
+                elif (root.val == root.left.val - 1):
+                    inr = left[0] + 1
+
+            if root.right:
+                right = longest_path(root.right)
+                if (root.val == root.right.val + 1):
+                    dcr = max(dcr, right[1] + 1)
+                elif (root.val == root.right.val - 1):
+                    inr = max(inr, right[0] + 1)
+
+            maxval = max(maxval, dcr + inr - 1)
+            return [inr, dcr]
+
+        maxval = 0
+        longest_path(root)
+        return maxval
+```
+
+```java
+class Solution {
+    int maxval = 0;
+
+    public int longestConsecutive(TreeNode root) {
+        longestPath(root);
+        return maxval;
+    }
+
+    public int[] longestPath(TreeNode root) {
+        if (root == null) {
+            return new int[] {0,0};
+        }
+
+        int inr = 1, dcr = 1;
+        if (root.left != null) {
+            int[] left = longestPath(root.left);
+            if (root.val == root.left.val + 1) {
+                dcr = left[1] + 1;
+            } else if (root.val == root.left.val - 1) {
+                inr = left[0] + 1;
+            }
+        }
+
+        if (root.right != null) {
+            int[] right = longestPath(root.right);
+            if (root.val == root.right.val + 1) {
+                dcr = Math.max(dcr, right[1] + 1);
+            } else if (root.val == root.right.val - 1) {
+                inr = Math.max(inr, right[0] + 1);
+            }
+        }
+
+        maxval = Math.max(maxval, dcr + inr - 1);
+        return new int[] {inr, dcr};
+    }
+}
+```
+
+```cpp
+class Solution {
+private:
+    int maxval = 0;
+
+    vector<int> longestPath(TreeNode* root) {
+        if (root == nullptr) {
+            return {0, 0};
+        }
+
+        int inr = 1, dcr = 1;
+
+        if (root->left != nullptr) {
+            vector<int> left = longestPath(root->left);
+            if (root->val == root->left->val + 1) {
+                dcr = left[1] + 1;
+            } else if (root->val == root->left->val - 1) {
+                inr = left[0] + 1;
+            }
+        }
+
+        if (root->right != nullptr) {
+            vector<int> right = longestPath(root->right);
+            if (root->val == root->right->val + 1) {
+                dcr = max(dcr, right[1] + 1);
+            } else if (root->val == root->right->val - 1) {
+                inr = max(inr, right[0] + 1);
+            }
+        }
+
+        maxval = max(maxval, dcr + inr - 1);
+
+        return {inr, dcr};
+    }
+
+public:
+    int longestConsecutive(TreeNode* root) {
+        longestPath(root);
+        return maxval;
+    }
+};
+```
+
+```javascript
+class Solution {
+    constructor() {
+        this.maxval = 0;
+    }
+
+    /**
+     * @param {TreeNode} root
+     * @return {number}
+     */
+    longestConsecutive(root) {
+        this.maxval = 0;
+        this.longestPath(root);
+        return this.maxval;
+    }
+
+    /**
+     * @param {TreeNode} root
+     * @return {number[]}
+     */
+    longestPath(root) {
+        if (root === null) {
+            return [0, 0];
+        }
+
+        let inr = 1, dcr = 1;
+
+        if (root.left !== null) {
+            let left = this.longestPath(root.left);
+            if (root.val === root.left.val + 1) {
+                dcr = left[1] + 1;
+            } else if (root.val === root.left.val - 1) {
+                inr = left[0] + 1;
+            }
+        }
+
+        if (root.right !== null) {
+            let right = this.longestPath(root.right);
+            if (root.val === root.right.val + 1) {
+                dcr = Math.max(dcr, right[1] + 1);
+            } else if (root.val === root.right.val - 1) {
+                inr = Math.max(inr, right[0] + 1);
+            }
+        }
+
+        this.maxval = Math.max(this.maxval, dcr + inr - 1);
+
+        return [inr, dcr];
+    }
+}
+```
+
+::tabs-end
+
+### Time & Space Complexity
+
+- Time complexity: $O(n)$
+- Space complexity: $O(n)$
+
+>  Where $n$ is the number of nodes in the input tree