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