最近公共祖先I

描述：

給定一棵二叉樹，找到兩個(gè)節(jié)點(diǎn)的最近公共父節(jié)點(diǎn) (LCA)。最近公共祖先是兩個(gè)節(jié)點(diǎn)的公共的祖先節(jié)點(diǎn)且具有最大深度。

! 注意事項(xiàng) : 假設(shè)給出的兩個(gè)節(jié)點(diǎn)都在樹中存在

樣例 ：

對于下面這棵二叉樹

  4
 / \
3   7
   / \
  5   6
輸出結(jié)果：
LCA(3, 5) = 4
LCA(5, 6) = 7
LCA(6, 7) = 7

代碼實(shí)現(xiàn)：

/**
 * 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 || root == A || root == B) {
            return root;
        }
        //divide and conquer
        TreeNode left = lowestCommonAncestor(root.left, A, B);
        TreeNode right = lowestCommonAncestor(root.right, A, B);
        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        if (left == null) {
            return right;
        }
        return null;
    }
}

最近公共祖先II

描述：

給一棵二叉樹和二叉樹中的兩個(gè)節(jié)點(diǎn)，找到這兩個(gè)節(jié)點(diǎn)的最近公共祖先LCA。
兩個(gè)節(jié)點(diǎn)的最近公共祖先，是指兩個(gè)節(jié)點(diǎn)的所有父親節(jié)點(diǎn)中（包括這兩個(gè)節(jié)點(diǎn)），離這兩個(gè)節(jié)點(diǎn)最近的公共的節(jié)點(diǎn)。每個(gè)節(jié)點(diǎn)除了左右兒子指針以外，還包含一個(gè)父親指針parent，指向自己的父親。

樣例：

對于下面的這棵二叉樹

  4
 / \
3   7
   / \
  5   6
輸出結(jié)果：
LCA(3, 5) = 4
LCA(5, 6) = 7
LCA(6, 7) = 7

代碼實(shí)現(xiàn)：

/**
 * 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;

    }
}

最近公共祖先 III

描述：

給一棵二叉樹和二叉樹中的兩個(gè)節(jié)點(diǎn)，找到這兩個(gè)節(jié)點(diǎn)的最近公共祖先 LCA。
兩個(gè)節(jié)點(diǎn)的最近公共祖先，是指兩個(gè)節(jié)點(diǎn)的所有父親節(jié)點(diǎn)中（包括這兩個(gè)節(jié)點(diǎn)），離這兩個(gè)節(jié)點(diǎn)最近的公共的節(jié)點(diǎn)。返回 null 如果兩個(gè)節(jié)點(diǎn)在這棵樹上不存在最近公共祖先的話。

樣例：

給出下面這棵樹：

  4
 / \
3   7
   / \
  5   6

輸出結(jié)果：
LCA(3, 5) = 4
LCA(5, 6) = 7
LCA(6, 7) = 7
LCA(5, 8) = null

代碼實(shí)現(xiàn)：

/**
 * 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;
 *     }
 * }
 */
 class ResultType {
    public boolean a_exist, b_exist;
    public TreeNode node;
    ResultType(boolean a, boolean b, TreeNode n) {
        a_exist = a;
        b_exist = b;
        node = n;
    }
}
public class Solution {
    /**
     * @param root The root of the binary tree.
     * @param A and B two nodes
     * @return: Return the LCA of the two nodes.
     */
    public TreeNode lowestCommonAncestor3(TreeNode root, TreeNode A, TreeNode B) {
        // write your code here
         // write your code here
        ResultType rt = helper(root, A, B);
        if (rt.a_exist && rt.b_exist)
            return rt.node;
        else
            return null;
    }

    public ResultType helper(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null)
            return new ResultType(false, false, null);
            
        ResultType left_rt = helper(root.left, A, B);
        ResultType right_rt = helper(root.right, A, B);
        
        boolean a_exist = left_rt.a_exist || right_rt.a_exist || root == A;
        boolean b_exist = left_rt.b_exist || right_rt.b_exist || root == B;
        
        if (root == A || root == B)
            return new ResultType(a_exist, b_exist, root);

        if (left_rt.node != null && right_rt.node != null) 
            return new ResultType(a_exist, b_exist, root);
        if (left_rt.node != null)
            return new ResultType(a_exist, b_exist, left_rt.node);
        if (right_rt.node != null)
            return new ResultType(a_exist, b_exist, right_rt.node);

        return new ResultType(a_exist, b_exist, null);
    }
}