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。

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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

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分析：最近公共祖先，比较常见的题了，先求出从顶点到A,B两的路径，再根据路径判断即可。

代码：

/**
 * 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
        vector<TreeNode*> APath;
        vector<TreeNode*> BPath;
        vector<TreeNode*> cur;
        dfs(cur,root,A,B,APath,BPath);
        TreeNode* ret = root;
        for(int i=0;i<min(APath.size(),BPath.size());i++)
        {
            if(APath[i]==BPath[i])
            {
                ret = APath[i];
            }
            else
                break;
        }
        return ret;
    }
    void dfs(vector<TreeNode*> cur,TreeNode*root,TreeNode *A, TreeNode *B,vector<TreeNode*> &APath,vector<TreeNode*> &BPath)
        {
            cur.push_back(root);
            if(root==A)
                APath = cur;
            if(root==B)
                BPath = cur;
            if(root->left)
                dfs(cur,root->left,A,B,APath,BPath);
            if(root->right)
                dfs(cur,root->right,A,B,APath,BPath);
        }
};