LeetCode2096 从二叉树一个节点到另一个节点每一步的方向

LeetCode2096 从二叉树一个节点到另一个节点每一步的方向

最近公共祖先的变形题.

# 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 getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:

        father, s, t = {}, None, None
        def dfs(cur: TreeNode) -> None:

            nonlocal s, t

            if cur.val == startValue: s = cur
            if cur.val == destValue: t = cur

            if cur.left:
                father[cur.left] = cur
                dfs(cur.left)
            if cur.right:
                father[cur.right] = cur
                dfs(cur.right)
        
        dfs(root)

        def get_path(cur: TreeNode) -> List[str]:
            
            ret = []
            while cur != root:
                parent = father[cur]
                if cur == parent.left: ret.append('L')
                elif cur == parent.right: ret.append('R')
                cur = parent
            return ret[::-1]
        
        s_path, t_path = get_path(s), get_path(t)
        while s_path and t_path and s_path[0] == t_path[0]:
            s_path.pop(0), t_path.pop(0)
        
        ans = 'U' * len(s_path) + ''.join(t_path)
        return ans