Lowest Common Ancestor of Two Nodes in a Binary Tree

Reference：

http://blog.csdn.net/v_july_v/article/details/18312089

http://leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-ii.html

(1) Is the tree a BST or not?

BST的话，我们就能按照BST的定义思考了。当前遍历的node如果比我们要找的两个点都大，说明那两个点都在当前node的左边，所以这两个node的祖先肯定在当前node的左边，所以往左边找。反之，往右边找。如果这两个点一个大于当前node一个小于当前node，说明当前node就是LCA。如果这两个点一个是另一个的祖先，那么这个点就是LCA。

代码如下：

1     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
2         if(p.val < root.val && q.val < root.val)
3             return lowestCommonAncestor(root.left, p, q);
4         else if(p.val > root.val && q.val > root.val)
5             return lowestCommonAncestor(root.right, p, q);
6         else
7             return root;

8 }

（2）那如果不是BST呢？一般Binary Tree.

a. Tree结构中没有parent域。

递归找。

代码如下：

1 //Worst case O(n)
2 Node getLCA(Node root, Node node1, Node node2){
3         if(root == null)
4             return null
5         if(root == node1 || root == node2)
6             return root;
7
8         Node left = getLCA(root.left, node1, node2);
9         Node right = getLCA(root.right, node1, node2);
10
11         if(left != null && right != null)
12             return root;
13         else if(left != null)
14             return left;
15         else if(right != null)
16             return right;
17         else
18             return null;
19     }

b. 如果有parent域。

转化为两个linkedlist的交点问题。

代码如下：

1 int getHeight(Node p){
2         int height = 0;
3         while(p!=null){
4             height++;
5             p = p.parent;
6         }
7         return height;
8     }
9
10     void swap(int a, int b){
11         a = a + b;
12         b = a - b;
13         a = a - b;
14     }
15
16     Node getLCA(Node p, Node q){
17         int h1 = getHeight(p);
18         int h2 = getHeight(q);
19
20         //q is always deeper than p
21         if(h1 > h2){
22             swap(h1, h2);
23             swap(p, q);
24         }
25
26         int diff = h2 - h1;
27
28         for( int i = 0; i < diff; i++)
29             q = q.parent;
30
31         while(p!=null && q!=null){
32             //common node
33             if(p == q)
34                 return p;
35             p = p.parent;
36             q = q.parent;
37         }
38
39         return NULL; //p and q are not in the same tree
40     }