236.Lowest Common Ancestor of a Binary Tree

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]

        _______3______
       /              
    ___5__          ___1__
   /              /      
   6      _2       0       8
         /  
         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 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.

Solution1:

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        def judge(root,child):
            if not root or not child:
                return False
            if root==child:
                return True
            return judge(root.left,child) or judge(root.right,child)
        dic = collections.defaultdict(bool)
        if root is None:
            return None
        dic[root] = True
        dic[root.left] = judge(root.left,p) and judge(root.left,q)
        dic[root.right] = judge(root.right,p) and judge(root.right,q)
        if not dic[root.left] and not dic[root.right]:
            return root
        elif dic[root.left]:
            return self.lowestCommonAncestor(root.left,p,q)
        else:
            return self.lowestCommonAncestor(root.right,p,q)

29 / 31 test cases passed.
root是p和q的LCA,当且仅当p和q都是root的子节点或其本身，root.left和root.right都不是。

Solution2:

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root is None or root==p or root==q: #发现目标节点则通过返回值标记该子树发现了某个目标结点
            return root
        left = self.lowestCommonAncestor(root.left,p,q) #查看左子树中是否有目标结点，没有为null
        right = self.lowestCommonAncestor(root.right,p,q) #查看右子树是否有目标节点，没有为null
        if left is not None and right is not None: #都不为空，说明左右子树都有目标结点，则公共祖先就是本身 
            return root
        return left if left is not None else right #You'd better use 'left is not None' rather than 'not left'