给定一棵二叉树，找到两个节点的最近公共父节点(LCA)。
最近公共祖先是两个节点的公共的祖先节点且具有最大深度。
样例
对于下面这棵二叉树

 4
 / \
3   7
   / \
  5   6

LCA(3, 5) = 4
LCA(5, 6) = 7
LCA(6, 7) = 7
题目链接：http://www.jianshu.com/writer#/notebooks/7169481/notes/8544182/preview
从根节点开始寻找A和B，然后根据路径来判断即可：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of the binary search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        // write your code here
        if (root == NULL) return root;
        bool la = postorder(root->left,A);
        bool lb = postorder(root->left,B);
        bool ra = postorder(root->right,A);
        bool rb = postorder(root->right,B);
        if (la && lb) return lowestCommonAncestor(root->left,A,B);
        else if (ra && rb) return lowestCommonAncestor(root->right,A,B);
        else return root;
    }
    bool postorder(TreeNode *root,TreeNode *A) {
        if (root == NULL) return false;
        if (root == A) return true;
        return postorder(root->left,A) || postorder(root->right,A);
    }
};