Lowest Common Ancestor lintcode

Given the root and two nodes in a Binary Tree. Find the lowest common ancestor(LCA) of the two nodes.

The lowest common ancestor is the node with largest depth which is the ancestor of both nodes.

Example

        4

    /     \

  3         7

          /     \

        5         6

For 3 and 5, the LCA is 4.

For 5 and 6, the LCA is 7.

For 6 and 7, the LCA is 7.
这道题还是用分治法， 从最底下往上遍历，当找到一个所给node，向上传递node， 如果没找到就传递null。 上一层的parent会check自己的左右子树是否都有返回值，a.如果都有值那么这个node就是LCA， 向上传递这个node一直到根节点。b. 只有一个子树有值， 那么向上传递这个子树。c.若果左右都没有， 向上传递null

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param A and B: two nodes in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null){
            return null;
        }
        if (root == A || root == B){
            return root;//当找到一中一个node， 向上传递这个node
        }
        TreeNode left = lowestCommonAncestor(root.left, A, B);
        TreeNode right = lowestCommonAncestor(root.right, A, B);
        if (left != null && right != null){//check左右，如果左右分别包含两个node则这个root就是LCA， 向上传递这个node
            return root;
        } else if (left != null){//只有左边有node，向上传递此node
            return left;
        } else if (right != null){
            return right;
        } else{//左右边都没有node 传递null
            return null;
        }
    }
}