Paste_Image.png

My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null)
            return null;
        return findAncestor(root, p, q);
    }
    
    private TreeNode findAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if ((root.val >= p.val && root.val <= q.val) || (root.val <= p.val && root.val >= q.val))
            return root;
        if (root.val < p.val && root.val < q.val)
            return findAncestor(root.right, p, q);
        else if (root.val > p.val && root.val > q.val)
            return findAncestor(root.left, p, q);
        return null;
    }
}

My test result:

Paste_Image.png

這道題目就比較簡單了。他和上面那道題的區(qū)別是什么。
如果p或者q == root，那么，他們的祖先一定是root。因為沒找到的那個，一定在他的左右子樹上。而在上道題目中并不一定。
相當于兩個點pq同時進入這棵樹。如果發(fā)現(xiàn)root的值正好大于等于其中一個然后小于等于另外一個，那么，他就一定是最小祖先。
否則，pq只能同時小于或者大于root.val，那么，再選擇下一步遍歷的方向。
很類似于pre-order,先判斷root結點
**
總結： pre-order tree
**

Anyway, Good luck, Richardo!

My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null) {
            return null;
        }
        else if (p == q) {
            return p;
        }
        TreeNode curr = root;
        HashSet<TreeNode> tracker = new HashSet<TreeNode>();
        while (curr != null) {
            if (curr.val < p.val) {
                tracker.add(curr);
                curr = curr.right;
            }
            else if (curr.val > p.val) {
                tracker.add(curr);
                curr = curr.left;
            }
            else {
                tracker.add(curr);
                break;
            }
        }
        
        curr = root;
        TreeNode pre = null;
        
        while (curr != null) {
            if (!tracker.contains(curr)) {
                return pre;
            }
            else if (curr.val < q.val) {
                pre = curr;
                curr = curr.right;
            }
            else if (curr.val > q.val) {
                pre = curr;
                curr = curr.left;
            }
            else {
                return q;
            }
        }
        
        return null;
    }
}

我的做法代碼比較復雜。
然后看了 Discuss,發(fā)現(xiàn)有更加簡潔的做法。
both iteration and recursion
具體看
reference:
https://discuss.leetcode.com/topic/18381/my-java-solution

Anyway, Good luck, Richardo！ -- 08/28/2016

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode curr = root;
        while (curr != null) {
            if (p.val <= curr.val && curr.val <= q.val) {
                return curr;
            }
            else if (q.val <= curr.val && curr.val <= p.val) {
                return curr;
            }
            else if (p.val < curr.val && q.val < curr.val) {
                curr = curr.left;
            }
            else {
                curr = curr.right;
            }
        }
        return null;
    }
}

空間復雜度是 O(1)

現(xiàn)在的狀態(tài)出奇的差。跟女朋友關系飄忽不定。
突然發(fā)現(xiàn)她，才是我力量的源泉。沒有了她，我完全沒有斗志了。
其實現(xiàn)在準備的也差不多了，過去真的就是各安天命。
整個浪費的時間并不多，也不可能因為浪費這點時間而改變結果。

把面經好好鞏固下，加油吧！

Anyway, Good luck, Richardo! -- 10/23/2016