2096. Step-By-Step Directions From a Binary Tree Node to Another

# Time:  O(n)
# Space: O(h)

# Definition for a binary tree node.
class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


class Solution(object):
    def getDirections(self, root, startValue, destValue):
        """
        :type root: Optional[TreeNode]
        :type startValue: int
        :type destValue: int
        :rtype: str
        """
        def iter_dfs(root, val):
            path = []
            stk = [(1, (root,))]
            while stk:
                step, args = stk.pop()
                if step == 1:
                    node = args[0]
                    if node.val == val:
                        path.reverse()
                        return path
                    for i, child in enumerate((node.left, node.right)):
                        if not child:
                            continue
                        stk.append((3, None))
                        stk.append((1, (child,)))
                        stk.append((2, ("LR"[i],)))
                elif step == 2:
                    path.append(args[0])
                elif step == 3:
                    path.pop()
            return []
    
        src = iter_dfs(root, startValue)
        dst = iter_dfs(root, destValue)
        while len(src) and len(dst) and src[-1] == dst[-1]:
            src.pop()
            dst.pop()
        dst.reverse()
        return "".join(['U']*len(src) + dst)


# Time:  O(n)
# Space: O(h)
class Solution2(object):
    def getDirections(self, root, startValue, destValue):
        """
        :type root: Optional[TreeNode]
        :type startValue: int
        :type destValue: int
        :rtype: str
        """
        def dfs(node, val, path):
            if node.val == val:
                return True
            if node.left and dfs(node.left, val, path):
                path.append('L')
            elif node.right and dfs(node.right, val, path):
                path.append('R')
            return path

        src, dst = [], []
        dfs(root, startValue, src)
        dfs(root, destValue, dst)
        while len(src) and len(dst) and src[-1] == dst[-1]:
            src.pop()
            dst.pop()
        dst.reverse()
        return "".join(['U']*len(src) + dst)

Solution from kamyu104/LeetCode-Solutions · MIT