Note

这道题和一不同，给了一个Node向上回溯的可能，所以不需要recersive的找。因为之前的那个题不能回头，所以必须先到最下面（或者找的A和B）。这道题我们只需要把A和B的path记住就可以了，然后比较path中从root到A或者B，一直到开始不一样的时候停止，那个最后一个一样的就是LCA。

/** * Definition of ParentTreeNode: *  * class ParentTreeNode { *     public ParentTreeNode parent, left, right; * } */public class Solution {    /**     * @param root: The root of the tree     * @param A, B: Two node in the tree     * @return: The lowest common ancestor of A and B     */    public ParentTreeNode lowestCommonAncestorII(ParentTreeNode root,                                                 ParentTreeNode A,                                                 ParentTreeNode B) {        // Write your code here        List
     
       pathA = findPath2Root(A);        List
      
        pathB = findPath2Root(B);                int indexA = pathA.size() - 1;        int indexB = pathB.size() - 1;                ParentTreeNode rst = null;        while (indexA >= 0 && indexB >= 0) {            if (pathA.get(indexA) != pathB.get(indexB)) {                break;            }            rst = pathA.get(indexA);            indexA--;            indexB--;        }        return rst;    }        private List
       
         findPath2Root(ParentTreeNode node) {        List
        
          path = new ArrayList
         
          ();        while (node != null) {            path.add(node);            node = node.parent;        }        return path;    } }