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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.

• /**
* 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.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);
}
}

• // 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;
}
};

• # 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


• /**
* 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
}