Problem description:

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 search tree: root = [3,5,1,6,2,0,8,null,null,7,4]

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     _______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.

Solution:

Recursively search left and right subtree.
If can find a root->val == p or q, then this root is at least contains one of the node’s ancestor.

So we find in left subtree and right subtree, if both left and right can find a value, then current root is the ancestor.

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

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root == p or root == q:
            return root
        left = right = None
        if root.left:
            left = self.lowestCommonAncestor(root.left, p, q)
        if root.right:
            right = self.lowestCommonAncestor(root.right, p, q)
        
        if left and right:
            return root
        else:
            return left or right
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root) return root;
        if(root->val == p->val ||root->val == q->val) return root;
        
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        
        if(left && right) return root;
        if(!left) return right;
        if(!right) return left;
    }
};

time complexity: $O(n)$
space complexity: $O(logn)$

reference:
https://goo.gl/NJbDFm
https://goo.gl/7BNspr