Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the : “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.

这道求二叉树的最小共同父节点的题是之前那道的Follow Up。跟之前那题不同的地方是，这道题是普通是二叉树，不是二叉搜索树，所以就不能利用其特有的性质，所以我们只能在二叉树中来搜索p和q，然后从路径中找到最后一个相同的节点即为父节点，我们可以用递归来实现，写法很简洁，代码如下：

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {       if (!root || p == root || q == root) return root;       TreeNode *left = lowestCommonAncestor(root->left, p, q);       TreeNode *right = lowestCommonAncestor(root->right, p , q);       if (left && right) return root;       return left ? left : right;    }};

上述代码可以进行优化一下，在找完左子树的共同父节点时如果结果存在，且不是p或q，那么不用再找右子树了，直接返回这个结果即可，同理，对找完右子树的结果做同样处理，参见代码如下：

class Solution {public:    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {       if (!root || p == root || q == root) return root;       TreeNode *left = lowestCommonAncestor(root->left, p, q);       if (left && left != p && left != q) return left;       TreeNode *right = lowestCommonAncestor(root->right, p , q);       if (right && right != p && right != q) return right;       if (left && right) return root;       return left ? left : right;    }};

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