Description#
You are given the root
of a binary tree.
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, move to the right child of the current node; otherwise, move to the left child.
- Change the direction from right to left or from left to right.
- Repeat the second and third steps 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:
![](https://spcdn.pages.dev/leetcode/problems/1372.Longest%20ZigZag%20Path%20in%20a%20Binary%20Tree/images/sample_1_1702.png)
Input: root = [1,null,1,1,1,null,null,1,1,null,1,null,null,null,1]
Output: 3
Explanation: Longest ZigZag path in blue nodes (right -> left -> right).
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/1372.Longest%20ZigZag%20Path%20in%20a%20Binary%20Tree/images/sample_2_1702.png)
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:
- The number of nodes in the tree is in the range
[1, 5 * 104]
. 1 <= Node.val <= 100
Solutions#
Solution 1#
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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.
* 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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| /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans = 0;
int longestZigZag(TreeNode* root) {
dfs(root, 0, 0);
return ans;
}
void dfs(TreeNode* root, int l, int r) {
if (!root) return;
ans = max(ans, max(l, r));
dfs(root->left, r + 1, 0);
dfs(root->right, 0, l + 1);
}
};
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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
}
|