［LeetCode］Lowest Common Ancestor of a Binary Tree

解题思路：
肯定要搜索这颗树。我这里有两个searchTree方法；
1，当p和q都没有被找到时，调用 int searchTree(TreeNode *root, TreeNode* p, TreeNode* q, TreeNode* &LCA)
2，当p和q有一个 == 该node，则调用int searchTree(TreeNode *root, TreeNode* p)
3，返回值为1，该子树匹配到一个node，返回值为2，该子树已经找到了LCA，返回值为0，没收获

注意一点：判断两个node是否相等，用 root == q 不能用root->val == q->val ；因为val有重复

/**
* 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) {
TreeNode *LCA = NULL;
searchTree(root, p, q, LCA);

return LCA;
}

int searchTree(TreeNode *root, TreeNode* p, TreeNode* q, TreeNode* &LCA){
if (root == NULL) return 0;
int ret = 0;

if (root == p){
ret ++;
if (searchTree(root->left, q) || searchTree(root->right, q) ){
LCA = root;
ret++;
}
}else if (root == q){
ret ++;
if (searchTree(root->left, p) || searchTree(root->right, p)){
LCA = root;
ret++;
}
}else{
int r = searchTree(root->left, p, q, LCA);
if (r == 2) return r;

int l = searchTree(root->right, p, q, LCA);
if (l == 2) return l;

ret = l+r;
if (ret == 2){
LCA = root;
}
}
return ret;
}

int searchTree(TreeNode *root, TreeNode* p){
if (root == NULL) return 0;
if (root == p){
return 1;
}
return searchTree(root->left, p) || searchTree(root->right, p);
}
};