Description
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].
分析
题目的意思是：找出二叉树的最长zigzag链，这道题我看了一下提示，要用动态规划的思想，用isLeft标记当前的结点需要向左遍历还是向右遍历，为True的话，下一步可以保持向左，也可以向右，向左的话，step+1，向右的话step要置1，表示不能跟构成zigzag链了，需要从头开始计数；相反为False的话，下一步可以保持向右，也可以向左，向右的话，step+1，向左的话，step要置1，表示不能构成zigzag链了，需要从头开始计数。
代码
# 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 maxZigZag(self,root,isLeft,step):
if(root is None):
return
self.maxStep=max(self.maxStep,step)
if(isLeft):
self.maxZigZag(root.right,True,1)
self.maxZigZag(root.left,False,step+1)
else:
self.maxZigZag(root.right,True,step+1)
self.maxZigZag(root.left,False,1)
def longestZigZag(self, root: TreeNode) -> int:
self.maxStep=0
self.maxZigZag(root,True,0)
self.maxZigZag(root,False,0)
return self.maxStep
参考文献
[LeetCode] [Java/C++] DFS Solution with comment - O(N) - Clean code