Lowest Common Ancestor of a Binary Tree.md

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
 

Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the binary tree.

My code:

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

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        #First_build the tree layer id and pointer
        self.build_tree(root,0,None)
        final_node=self.search(p,q)
        return final_node
    def search(self,p,q):
        while q.layer>p.layer:
            q=q.root
        while q.layer<p.layer:
            p=p.root
        if p==q:
            return p
        p=p.root
        q=q.root
        result=self.search(p,q)
        return result
    def build_tree(self,root,layer,pointer):
        if root==None:
            return
        root.root=pointer
        root.layer=layer
        left=root.left
        right=root.right
        self.build_tree(left,layer+1,root)
        self.build_tree(right,layer+1,root)
        

Solution: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/