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Formatted question description: https://leetcode.ca/all/1372.html
1372. Longest ZigZag Path in a Binary Tree (Medium)
Given a binary tree root
, a ZigZag path for a binary tree is defined as follow:
- Choose any node in the binary tree and a direction (right or left).
- If the current direction is right then move to the right child of the current node otherwise move to the left child.
- Change the direction from right to left or right to left.
- Repeat the second and third step until you can't move in the tree.
Zigzag length is defined as the number of nodes visited - 1. (A single node has a length of 0).
Return the longest ZigZag path contained in that tree.
Example 1:
Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1,null,1] Output: 3 Explanation: Longest ZigZag path in blue nodes (right -> left -> right).
Example 2:
Input: root = [1,1,1,null,1,null,null,1,1,null,1] Output: 4 Explanation: Longest ZigZag path in blue nodes (left -> right -> left -> right).
Example 3:
Input: root = [1] Output: 0
Constraints:
- Each tree has at most
50000
nodes.. - Each node's value is between
[1, 100]
.
Related Topics:
Dynamic Programming, Tree
Solution 1.
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int longestZigZag(TreeNode root) { Map<TreeNode, int[]> map = new HashMap<TreeNode, int[]>(); map.put(root, new int[]{0, 0}); int maxLength = 0; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); int[] lengths = map.get(node); TreeNode left = node.left, right = node.right; if (left != null) { int[] leftLengths = new int[2]; leftLengths[0] = lengths[1] + 1; maxLength = Math.max(maxLength, leftLengths[0]); map.put(left, leftLengths); queue.offer(left); } if (right != null) { int[] rightLengths = new int[2]; rightLengths[1] = lengths[0] + 1; maxLength = Math.max(maxLength, rightLengths[1]); map.put(right, rightLengths); queue.offer(right); } } return maxLength; } } ############ /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { private int ans; public int longestZigZag(TreeNode root) { dfs(root, 0, 0); return ans; } private void dfs(TreeNode root, int l, int r) { if (root == null) { return; } ans = Math.max(ans, Math.max(l, r)); dfs(root.left, r + 1, 0); dfs(root.right, 0, l + 1); } }
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// OJ: https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree // Time: O(N) // Space: O(H) class Solution { int A = 0; pair<int, int> postorder(TreeNode* root) { if (!root || (!root->left && !root->right)) return { 0, 0 }; pair<int, int> ans; if (root->left) { auto left = postorder(root->left); ans.first = 1 + left.second; } if (root->right) { auto right = postorder(root->right); ans.second = 1 + right.first; } A = max({ A, ans.first, ans.second }); return ans; } public: int longestZigZag(TreeNode* root) { postorder(root); return A; } };
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# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def longestZigZag(self, root: TreeNode) -> int: def dfs(root, l, r): if root is None: return nonlocal ans ans = max(ans, l, r) dfs(root.left, r + 1, 0) dfs(root.right, 0, l + 1) ans = 0 dfs(root, 0, 0) return ans
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/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func longestZigZag(root *TreeNode) int { ans := 0 var dfs func(root *TreeNode, l, r int) dfs = func(root *TreeNode, l, r int) { if root == nil { return } ans = max(ans, max(l, r)) dfs(root.Left, r+1, 0) dfs(root.Right, 0, l+1) } dfs(root, 0, 0) return ans } func max(a, b int) int { if a > b { return a } return b }