lowest common ancestor of a binary tree medium

Author: weiy

Diff for: Tree/LowestCommonAncestorOfABinaryTree.py

@@ -0,0 +1,91 @@
+"""
+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.
+
+
+给一颗二叉树，找出某两个子节点的最小公共祖先。
+思路：
+
+用递归：
+   1. 用递归找到符合条件的子节点。 找到后的子节点会返回为一个具体的 TreeNode，找不到的话则是 None。
+   2. 之后判断 是不是两个都找到了，最先知道两个都找到的点即为最小公共祖先。
+   3. q为p子节点，或p为q子节点的情况：
+         由于是唯一的，所以出现这种情况一定有一边返回是None，所以返回不是None的一边即可。
+
+普通的二叉树要递归的话是这样：
+
+# do something
+
+if root.right:
+    right = recursive(root.right)
+if root.left:
+    left = recursive(root.left)
+
+# do something
+
+按照上面的思路加工一下即可。
+
+测试地址：
+https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
+
+"""
+# 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
+        """
+        if root.val == p.val or root.val == q.val:
+            return root
+        
+        right = None
+        left = None
+        
+        if root.right:
+            right = self.lowestCommonAncestor(root.right, p, q)
+        if root.left:
+            left = self.lowestCommonAncestor(root.left, p, q)
+        
+        if right and left:
+            return root
+        
+        if right:
+            return right
+        
+        if left:
+            return left