[LeetCode] Lowest Common Ancestor of a Binary Tree

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.

解题思路：

这道题的意思是求出二叉树任何两点的最低公共节点。与二分查找树的不同是，这个查找并没有规律，只能暴力法进行。

解法一：

递归法。倘若p, q都在root的左孩子树中，则令root=root->left。倘若p, q都在root的右孩子树种，则令root=root->right。否则，返回root。这种方法会有很多重复计算，产生超时错误。

/**
* 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;
}
if(isInTree(root->left, p) && isInTree(root->left, q)){
return lowestCommonAncestor(root->left, p, q);
}else if(isInTree(root->right, p) && isInTree(root->right, q)){
return lowestCommonAncestor(root->right, p, q);
}else{
return root;
}
}

bool isInTree(TreeNode* root, TreeNode* p){
if(root==NULL){
return false;
}
return root == p || isInTree(root->left, p) || isInTree(root->right, p);
}
};解法二：
可以先求出p，q到根节点的路径，然后找到路径中第一个相同的节点。

/**
* 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;
}
if(p==q){
return p;
}
vector<TreeNode*> pPath;
vector<TreeNode*> qPath;
getPath(root, p, pPath);
getPath(root, q, qPath);
int i = pPath.size() - 1;
int j = qPath.size() - 1;
while(i>=0 && j>=0){
if(pPath[i] == qPath[j]){
i--;
j--;
}else{
return pPath[i+1];
}
}

return pPath[i+1];
}

bool getPath(TreeNode* root, TreeNode* p, vector<TreeNode*>& pPath){
if(root==NULL){
return false;
}
if(root==p){
pPath.push_back(root);
return true;
}
if(getPath(root->left, p, pPath) || getPath(root->right, p, pPath)){
pPath.push_back(root);
return true;
}
return false;
}
};