lintcode474- Lowest Common Ancestor II- easy

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.

The node has an extra attribute parent which point to the father of itself. The root's parent is null.

Example

For the following binary tree:

  4
 / 
3   7
   / 
  5   6

LCA(3, 5) = 4

LCA(5, 6) = 7

LCA(6, 7) = 7

1.用hashset辅助做。思想是AB各自向上回溯，如果第一次碰到了一样的点那就是lca.所以可以先回溯a，都存到set里，再在回溯b的时候每次看访问的点在不在a的父亲们里面了。

2.用两个arraylist做。思想是把ab各自到root的路径全都打印出来。那么这个root如果倒过来看的话，前面一部分都会是一样的，第一个不一样的点的上面那个分岔点就是lca了。

1.我的代码

/**
 * Definition of ParentTreeNode:
 * 
 * class ParentTreeNode {
 *     public ParentTreeNode parent, left, right;
 * }
 */


public class Solution {
    /*
     * @param root: The root of the tree
     * @param A: node in the tree
     * @param B: 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
        
        ParentTreeNode result = null;
        Set<ParentTreeNode> set = new HashSet<ParentTreeNode>();
        
        ParentTreeNode aParent = A;
        ParentTreeNode bParent = B;
        
        while (aParent != null) {
            set.add(aParent);
            aParent = aParent.parent;
        }
        
        while (bParent != null) {
            if (set.contains(bParent)) {
                result = bParent;
                break;
            }
            bParent = bParent.parent;
        }
        
        return result;
        
    }
}

2.九章算法上的参考代码

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*/ 

/**
 * 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) {
        ArrayList<ParentTreeNode> pathA = getPath2Root(A);
        ArrayList<ParentTreeNode> pathB = getPath2Root(B);
        
        int indexA = pathA.size() - 1;
        int indexB = pathB.size() - 1;
        
        ParentTreeNode lowestAncestor = null;
        while (indexA >= 0 && indexB >= 0) {
            if (pathA.get(indexA) != pathB.get(indexB)) {
                break;
            }
            lowestAncestor = pathA.get(indexA);
            indexA--;
            indexB--;
        }
        
        return lowestAncestor;
    }
    
    private ArrayList<ParentTreeNode> getPath2Root(ParentTreeNode node) {
        ArrayList<ParentTreeNode> path = new ArrayList<>();
        while (node != null) {
            path.add(node);
            node = node.parent;
        }
        return path;
    }
}