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]
.
Solution: Recursion
For each node return
1. max ZigZag length if go left
2. max ZigZag length if go right
3. maz ZigZag length within the subtree
ZigZag(root):
ll, lr, lm = ZigZag(root.left)
rl, rr, rm = ZigZag(root.right)
return (lr+1, rl + 1, max(lr+1, rl+1, lm, rm))
Time complexity: O(n)
Space complexity: O(h)
C++
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// Author: Huahua class Solution { public: int longestZigZag(TreeNode* root) { return get<2>(ZigZag(root)); } // Returns {left, right, max} tuple<int, int, int> ZigZag(TreeNode* root) { if (!root) return {-1, -1, -1}; auto [ll, lr, lm] = ZigZag(root->left); auto [rl, rr, rm] = ZigZag(root->right); int l = lr + 1; int r = rl + 1; return {l, r, max({l, r, lm, rm})}; } }; |
Python3
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# Author: Huahua class Solution: def longestZigZag(self, root: TreeNode) -> int: def ZigZag(node): if not node: return (-1, -1, -1) ll, lr, lm = ZigZag(node.left) rl, rr, rm = ZigZag(node.right) return (lr + 1, rl + 1, max(lr + 1, rl + 1, lm, rm)) return ZigZag(root)[2] |
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