codingEskimo

摘要： Note: // 在root为根的二叉树中找A,B的LCA: // 如果找到了就返回这个LCA // 如果只碰到A，就返回A // 如果只碰到B，就返回B // 如果都没有，就返回null Just consider relationship between left, right and root 阅读全文

摘要： Note: It is easier to use divided conquer. As you can see, the question is just to add right to left's last. Here we care more about last then the top 阅读全文

摘要： Node: Quick Select: Avg O(N) Thought: put all the element less(more) than pivot to the first part, the rest to the second part. And Kth only falls to 阅读全文