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.
找最近的公共祖先，仔细想一想，其实l与r的最小的公共祖先c满足一定条件，那就是l以及r一定在c的左右分支上，不可能都是左或右分支的。否则一定就不是最近的公共祖先，
代码如下，用递归写出来还是比较简单易懂的。
/**
* 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; //无其他节点，直接返回
if (root == p || root == q) return root;
TreeNode * leftNode = lowestCommonAncestor(root->left, p, q);
TreeNode * rightNode = lowestCommonAncestor(root->right, p, q);
if (leftNode && rightNode) return root; //找到LCA，返回LCA
return leftNode ? leftNode : rightNode;
}
};
还有一种方法是遍历tree，然后找出到达p以及q分别的路径，找到路径之后，遍历两条路径，出现分叉的第一个点就是p与q的LCA。具体代码先不贴了 比较麻烦。
java版本的如下所示，方法相同：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root == null)
return null;
if(root == p || root == q)//切断路径，不再向下执行了
return root;
TreeNode leftNode = lowestCommonAncestor(root.left, p, q);
TreeNode rightNode = lowestCommonAncestor(root.right, p, q);
if(leftNode != null && rightNode != null)
return root;
if(leftNode!=null) return leftNode;
if(rightNode != null) return rightNode;
return null;
}
}