Lowest Common Ancestor of a Binary Tree

题目描述：

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.

解题思路：分别找到从树的根结点root到结点p和q的路径，然后两条路径的最后一个公共结点即为所要求的结点p和q的最小公共祖先
AC代码如下：
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 || p == NULL || q == NULL) return NULL;
vector<TreeNode*> path1;
vector<TreeNode*> path2;
getPath(root, p, path1);
getPath(root, q, path2);
TreeNode* ans = lastCommonNode(path1, path2);
return ans;
}
bool getPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& path){
if (root == p){
path.push_back(root);
return true;
}
bool findLeft = false;
bool findRight = false;
path.push_back(root);
if (root->left != NULL){
findLeft = getPath(root->left, p, path);
}
if (!findLeft && root->right != NULL){
findRight = getPath(root->right, p, path);
}
if (!findLeft && !findRight){
path.pop_back();
return false;
}
return true;
}
TreeNode* lastCommonNode(vector<TreeNode*>& path1, vector<TreeNode*>& path2){
vector<TreeNode*>::iterator it1 = path1.begin();
vector<TreeNode*>::iterator it2 = path2.begin();
TreeNode* ans = NULL;
while (it1 != path1.end() && it2 != path2.end()){
if (*it1 == *it2){
ans = *it1;
}
++it1;
++it2;
}
return ans;
}
};