数据结构与算法(1)支线任务4——Lowest Common Ancestor of a Binary Tree

题目如下：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______

       /              \

    ___5__          ___1__

   /      \        /      \

   6      _2       0       8

         /  \

         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

由于题目已经给定了数据结构和函数（见下方代码部分），并不能愉快地用指向双亲的指针，然而寻找p和q两个结点的公共祖先似乎又要从下方向上找。于是我们可以考虑用递归：要找的是第一个左右孩子都是p或q祖先的结点；或者本身是p，而且是q的祖先的结点。如果符合这两点，直接返回就好；如果不是任何一个祖先，忽略掉；如果是一个的祖先，可以通过递归将这个情况返回，实现从下向上找。

代码如下：

/**

 * 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) {

    if (root == NULL)

    {

        return NULL; //到了最下面，什么都没有，返回空

    }

    if (root == p || root == q)

    {

        return root; //找到了一个结点，返回该结点

    }

    //递归

    TreeNode *left = lowestCommonAncestor(root->left, p, q);

    TreeNode *right = lowestCommonAncestor(root->right, p, q);

    if (left && right)

    {

        return root; //待查找结点祖先分别是左右孩子，该结点为结果

    }

    else if (left)

    {

        return left; //只是一个的祖先，将结果“上传”

    }

    return right;

    }

};