今天芝加哥虽然很冷，但阳光明媚，是个刷题的好日子。

昨天晚上跟孔神一番交流后感受到了巨大差距，无论是java基础，算法基础，system design基础，项目经验还是努力程度，差距都是全方位的，想到自己一直以来都缺乏毅力，不能做到有始有终，也经常被老爸和suki诟病。

所以从今天起我准备每天都至少刷五道题，熟练后增加，立帖为证，自我激励，往者不可谏，来者犹可追。

祝愿自己能在接下来的几个月里找到满意的工作。

废话不多说，开始写题：

Stack：

155. min stack
     maintain two stacks, s1 is to store all push(), s2 is to store min values(every push() in one correspond to a min value in two)

private Deques1; 
 private Deques2;
 public MinStack() { 
     s1 = new LinkedList(); 
     s2 = new LinkedList();
    }

232. implement queue using stacks
     maintain move() to complete moving from s1 to s2.

public void move(){       
    if (s1.isEmpty()) {           
        return;       
    }       
    if(s2.isEmpty()) {           
        while (!s1.isEmpty()) {               
            s2.offerFirst(s1.pollFirst());           
        }       
    }   
}

150. Evaluate Reverse Polish Notation med

using a stack, when meeting string .equals() operators, pop the top two int in stack

note: string to int Integer.parseInt()
stack.offerFirst(Integer.parseInt(tk));

Binary Tree

110. Balanced Binary Tree
     recursively get every node's height, to the left and right subTrees should follow abs(left - right) < 1
     getHeight()
     subProblem: root.left & root.right 's height
     base case: root == null return 0
     isBalanced()
     if Math.abs(leftHeight, rightHeight) > 1 false
     we say, to every node
     subPro: isBalanced(root.left) && isBalanced(root.right)

111. Symmetric Tree
     need to prove that the left subtree and the right subtree are symmetric
     subproblem:
     isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left)
     corner case:

if (left == null || right == null) {
            return left == null && right == null;
        }
        if (left.val != right.val) {
            return false;
        }

100. Same Tree
     base case & subproblem

if (p.val != q.val) {
             return false;
        }
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);

98 valid BST
method1: inorder traverse the tree and when print new node, compare with the previous one.
method2: root.left subtree range (min, root)
root.right (root, max)

if (root.val <= min || root.val >= max) {
            return false;
        }
        return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);

Extra problem： Get Keys In Binary Search Tree In Given Range

在inorder in recursive way框架基础上加限制条件来剪枝

104. Maximum Depth of Binary Tree
     同110 利用height() 求出左右子树高度比较即可

105. Minimum Depth of Binary Tree
     当没有左子树或者右子树时， 得分开判断

if (root.left != null && root.right != null)
        return Math.min(minDepth(root.left), minDepth(root.right))+1;
    else
        return Math.max(minDepth(root.left), minDepth(root.right))+1;
    }