|
| 1 | +## 1. Brute Force (Time Limit Exceeded) |
| 2 | + |
| 3 | +### Time & Space Complexity |
| 4 | + |
| 5 | +- Time complexity: $O(n^3)$ |
| 6 | +- Space complexity: $O(n^3)$ |
| 7 | + |
| 8 | +> Where $n$ is the number of nodes in the input tree |
| 9 | +
|
| 10 | +--- |
| 11 | + |
| 12 | +## 2. Single traversal |
| 13 | + |
| 14 | +::tabs-start |
| 15 | + |
| 16 | +```python |
| 17 | +class Solution: |
| 18 | + def longestConsecutive(self, root: TreeNode) -> int: |
| 19 | + |
| 20 | + def longest_path(root: TreeNode) -> List[int]: |
| 21 | + nonlocal maxval |
| 22 | + |
| 23 | + if not root: |
| 24 | + return [0, 0] |
| 25 | + |
| 26 | + inr = dcr = 1 |
| 27 | + if root.left: |
| 28 | + left = longest_path(root.left) |
| 29 | + if (root.val == root.left.val + 1): |
| 30 | + dcr = left[1] + 1 |
| 31 | + elif (root.val == root.left.val - 1): |
| 32 | + inr = left[0] + 1 |
| 33 | + |
| 34 | + if root.right: |
| 35 | + right = longest_path(root.right) |
| 36 | + if (root.val == root.right.val + 1): |
| 37 | + dcr = max(dcr, right[1] + 1) |
| 38 | + elif (root.val == root.right.val - 1): |
| 39 | + inr = max(inr, right[0] + 1) |
| 40 | + |
| 41 | + maxval = max(maxval, dcr + inr - 1) |
| 42 | + return [inr, dcr] |
| 43 | + |
| 44 | + maxval = 0 |
| 45 | + longest_path(root) |
| 46 | + return maxval |
| 47 | +``` |
| 48 | + |
| 49 | +```java |
| 50 | +class Solution { |
| 51 | + int maxval = 0; |
| 52 | + |
| 53 | + public int longestConsecutive(TreeNode root) { |
| 54 | + longestPath(root); |
| 55 | + return maxval; |
| 56 | + } |
| 57 | + |
| 58 | + public int[] longestPath(TreeNode root) { |
| 59 | + if (root == null) { |
| 60 | + return new int[] {0,0}; |
| 61 | + } |
| 62 | + |
| 63 | + int inr = 1, dcr = 1; |
| 64 | + if (root.left != null) { |
| 65 | + int[] left = longestPath(root.left); |
| 66 | + if (root.val == root.left.val + 1) { |
| 67 | + dcr = left[1] + 1; |
| 68 | + } else if (root.val == root.left.val - 1) { |
| 69 | + inr = left[0] + 1; |
| 70 | + } |
| 71 | + } |
| 72 | + |
| 73 | + if (root.right != null) { |
| 74 | + int[] right = longestPath(root.right); |
| 75 | + if (root.val == root.right.val + 1) { |
| 76 | + dcr = Math.max(dcr, right[1] + 1); |
| 77 | + } else if (root.val == root.right.val - 1) { |
| 78 | + inr = Math.max(inr, right[0] + 1); |
| 79 | + } |
| 80 | + } |
| 81 | + |
| 82 | + maxval = Math.max(maxval, dcr + inr - 1); |
| 83 | + return new int[] {inr, dcr}; |
| 84 | + } |
| 85 | +} |
| 86 | +``` |
| 87 | + |
| 88 | +```cpp |
| 89 | +class Solution { |
| 90 | +private: |
| 91 | + int maxval = 0; |
| 92 | + |
| 93 | + vector<int> longestPath(TreeNode* root) { |
| 94 | + if (root == nullptr) { |
| 95 | + return {0, 0}; |
| 96 | + } |
| 97 | + |
| 98 | + int inr = 1, dcr = 1; |
| 99 | + |
| 100 | + if (root->left != nullptr) { |
| 101 | + vector<int> left = longestPath(root->left); |
| 102 | + if (root->val == root->left->val + 1) { |
| 103 | + dcr = left[1] + 1; |
| 104 | + } else if (root->val == root->left->val - 1) { |
| 105 | + inr = left[0] + 1; |
| 106 | + } |
| 107 | + } |
| 108 | + |
| 109 | + if (root->right != nullptr) { |
| 110 | + vector<int> right = longestPath(root->right); |
| 111 | + if (root->val == root->right->val + 1) { |
| 112 | + dcr = max(dcr, right[1] + 1); |
| 113 | + } else if (root->val == root->right->val - 1) { |
| 114 | + inr = max(inr, right[0] + 1); |
| 115 | + } |
| 116 | + } |
| 117 | + |
| 118 | + maxval = max(maxval, dcr + inr - 1); |
| 119 | + |
| 120 | + return {inr, dcr}; |
| 121 | + } |
| 122 | + |
| 123 | +public: |
| 124 | + int longestConsecutive(TreeNode* root) { |
| 125 | + longestPath(root); |
| 126 | + return maxval; |
| 127 | + } |
| 128 | +}; |
| 129 | +``` |
| 130 | +
|
| 131 | +```javascript |
| 132 | +class Solution { |
| 133 | + constructor() { |
| 134 | + this.maxval = 0; |
| 135 | + } |
| 136 | + |
| 137 | + /** |
| 138 | + * @param {TreeNode} root |
| 139 | + * @return {number} |
| 140 | + */ |
| 141 | + longestConsecutive(root) { |
| 142 | + this.maxval = 0; |
| 143 | + this.longestPath(root); |
| 144 | + return this.maxval; |
| 145 | + } |
| 146 | + |
| 147 | + /** |
| 148 | + * @param {TreeNode} root |
| 149 | + * @return {number[]} |
| 150 | + */ |
| 151 | + longestPath(root) { |
| 152 | + if (root === null) { |
| 153 | + return [0, 0]; |
| 154 | + } |
| 155 | + |
| 156 | + let inr = 1, dcr = 1; |
| 157 | + |
| 158 | + if (root.left !== null) { |
| 159 | + let left = this.longestPath(root.left); |
| 160 | + if (root.val === root.left.val + 1) { |
| 161 | + dcr = left[1] + 1; |
| 162 | + } else if (root.val === root.left.val - 1) { |
| 163 | + inr = left[0] + 1; |
| 164 | + } |
| 165 | + } |
| 166 | + |
| 167 | + if (root.right !== null) { |
| 168 | + let right = this.longestPath(root.right); |
| 169 | + if (root.val === root.right.val + 1) { |
| 170 | + dcr = Math.max(dcr, right[1] + 1); |
| 171 | + } else if (root.val === root.right.val - 1) { |
| 172 | + inr = Math.max(inr, right[0] + 1); |
| 173 | + } |
| 174 | + } |
| 175 | + |
| 176 | + this.maxval = Math.max(this.maxval, dcr + inr - 1); |
| 177 | + |
| 178 | + return [inr, dcr]; |
| 179 | + } |
| 180 | +} |
| 181 | +``` |
| 182 | + |
| 183 | +::tabs-end |
| 184 | + |
| 185 | +### Time & Space Complexity |
| 186 | + |
| 187 | +- Time complexity: $O(n)$ |
| 188 | +- Space complexity: $O(n)$ |
| 189 | + |
| 190 | +> Where $n$ is the number of nodes in the input tree |
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