1372. Longest ZigZag Path in a Binary Tree

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].

  class Solution {
      int maxStep = 0;
      public int longestZigZag(TreeNode root) {
          dfs(root, true, 0);
          dfs(root, false, 0);
          return maxStep;
      }
      private void dfs(TreeNode root, boolean isLeft, int step) {
          if (root == null) return;
          maxStep = Math.max(maxStep, step); // update max step sofar
          if (isLeft) {
              dfs(root.left, false, step + 1); // keep going from root to left
              dfs(root.right, true, 1); // restart going from root to right
          } else {
              dfs(root.right, true, step + 1); // keep going from root to right
              dfs(root.left, false, 1); // restart going from root to left
          }
      }
  }

  噫 DFS好，用一个boolean控制向左向右，二叉树还是太菜了我

• https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree/discuss/534418/Java-DFS-Solution-with-comment-O(N)-Clean-code