题目：
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.
分析：
求二叉树节点最低公共祖先的题目，应该做到下面两点：
1、时间复杂度O(n)，仅仅需遍历一遍二叉树。
2、要推断输入的两个节点是否在二叉树中。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct node
{
TreeNode* res;
bool bp,bq;
node(TreeNode* n,bool x,bool y):res(n),bp(x),bq(y){}
};
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
if(!root || !p || !q)
return nullptr;
return lowestCommonAncestor_core(root,p,q).res;
}
node lowestCommonAncestor_core(TreeNode* root, TreeNode* p, TreeNode* q)
{
if(root==nullptr)
return node(nullptr,false,false);
if(root==p)
{
return hasnode(root,q)?node(root,true,true):node(root,true,false);
}
if(root==q)
{
return hasnode(root,p)?node(root,true,true):node(root,false,true);
}
if(root->left)
{
auto r=lowestCommonAncestor_core(root->left,p,q);
if(r.bp && r.bq)
return r;
else if(!r.bp && r.bq)
{
if(hasnode(root->right,p))
return node(root,true,true);
else
return node(nullptr,false,true);
}
else if(r.bp && !r.bq)
{
if(hasnode(root->right,q))
return node(root,true,true);
else
return node(nullptr,true,false);
}
}
return lowestCommonAncestor_core(root->right,p,q);
}
bool hasnode(TreeNode* root, TreeNode* p)
{
if(root==p)
return true;
if(root==nullptr)
return false;
return hasnode(root->left,p) || hasnode(root->right,p);
}
};