Level：
Medium
题目描述：
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.
思路分析：
在一个二叉树中寻找两个节点的最小公共祖先，如果根不为空时，不断从其左右子树搜索，如果两个节点都在左子树，就在左子树递归查找，如果两个节点都在右子树，就在右子树递归查找，如果一个节点在左子树，一个节点在右字数，那么当前节点就是最小公共祖先。
代码：
public class TreeNode{
int val;
TreeNode left;
TreeNode right;
TreeNode(int x){
val=x;
}
}
public class Solution{
public TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q){
if(root==null||root==p||root==q)
return root;
//查看左子树有没有目标节点，没有就为null
TreeNode left=lowestCommonAncestor( root.left, p,q);
//查看右子树有没有目标节点，没有就为null
TreeNode right=lowestCommonAncestor( root.right, p,q);
//都不为空说明左右子树都有目标节点，那么当前节点就是最小祖先
if(left!=null&&right!=null)
return root;
//左子树为空就去右子树找，否则去左子树找
return left!=null?left:right;
}
}