Similar with finding lowest common ancestor of two nodes, we can extend this function to N nodes.

The method **buildTree** in the solution is in order to building a tree instantly by given strings.

# Definition for a binary tree node. class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None class Solution: def buildTree(self, inorder, postorder): if not inorder or not postorder: return None while postorder: if postorder[-1] not in inorder: postorder.pop() else: break if len(postorder) == 1: return TreeNode(postorder[0]) else: node = TreeNode(postorder[-1]) in_index = inorder.index(postorder[-1]) node.left = self.buildTree(inorder[:in_index], postorder[:-1]) node.right = self.buildTree(inorder[in_index + 1:], postorder[:-1]) return node def find(self, root, nodes): if not root: return None else: if root.val in nodes: return root left = self.find(root.left, nodes) right = self.find(root.right, nodes) if left and right: return root return right if not left else left if __name__ == "__main__": obj = Solution() root = obj.buildTree([6, 5, 7, 2, 4, 3, 0, 1, 8], [6, 7, 4, 2, 5, 0, 8, 1, 3]) nodes = [0, 1, 8] result = obj.find(root, nodes) print(result.val)