Note

這道題和一不同，給了一個Node向上回溯的可能，所以不需要recersive的找。因為之前的那個題不能回頭，所以必須先到最下面（或者找的A和B）。這道題我們只需要把A和B的path記住就可以了，然後比較path中從root到A或者B，一直到開始不一樣的時候停止，那個最後一個一樣的就是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<ParentTreeNode> pathA = findPath2Root(A);
        List<ParentTreeNode> 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<ParentTreeNode> findPath2Root(ParentTreeNode node) {
        List<ParentTreeNode> path = new ArrayList<ParentTreeNode>();
        while (node != null) {
            path.add(node);
            node = node.parent;
        }
        return path;
    } 
}

Lowest Common Ancestor II