Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
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 binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
_______6______ / \ ___2__ ___8__ / \ / \ 0 _4 7 9 / \ 3 5
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Note:
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All of the nodes' values will be unique.
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p and q are different and both values will exist in the BST.
给定二叉搜索树（BST），找到BST中两个给定节点的最低共同祖先（LCA）。
根据维基百科上LCA的定义：“最低共同祖先在两个节点p和q之间定义为T中的最低节点，其中p和q都是后代（我们允许节点成为其自身的后代）。 ”
给定二叉搜索树：root = [6,2,8,0,4,7,9，null，null，3,5]
思路
1.如果在左右子树中能分别找到p和q，那么它们的共同祖先就是root；
2.如果在左子树能找到p和q，那么左子树就是共同祖先；
3.如果在右子树能找到p和q，那么右子树就是共同祖先；
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
if(!root || root == p || root == q) return root;//当前结点为NULL，返回NULL；当前结点为p或q，返回root；
struct TreeNode *l = lowestCommonAncestor(root->left, p, q);//递归遍历左子树，找p和q，如果先找到p，返回p；先找到q就返回q；
struct TreeNode *r = lowestCommonAncestor(root->right, p, q);//递归遍历右子树；
if(l && r) return root;//如果左右子树都能找到p和q，那么root就是最低共同祖先，如Eg1所示；
if(!l) return r;//如果左子树中找不到p和q，那么在右子树中先找到的p或q就是LCA；
if(!r) return l;//如果右子树中找不到p和q，那么在左子树中先找到的p或q就是LCA；
return root;
}