VMware coding Challenge - Binary Tree Height


VMware coding Challenge - Binary Tree Height
思路:这道题要观察,举个例子,1 2 * * 3 * 4 5 * * 6 7 * 8 * *, 用Stack,先序遍历,遇到数字就入栈,如果遇到 * *,说明栈顶节点是叶子节点,一条根到叶子的路径这时候就存在于栈之中,只要计算栈的size(),就知道当前这条路径的深度,树的height就是这些深度的最大值。

空格的引入增添了不少判断上的麻烦, 比如: ab * *, 这棵树的height是1 而不是2. 在遍历节点的时候需要注意取空格之前的所有元素一起看

第三遍办法:倒序读数组,用stack存。遇到“*”存入0,遇到数字,pop两次,取最大值+1,再push入栈
 2     public int treeHeight(String preorder) {
 3         if (preorder==null || preorder.length()==0) return -1;
 4         preorder.trim();
 5         if (preorder.length() == 0) return -1;
 6         String[] strs = preorder.split(" ");
 7         int len = strs.length;
 8         Stack<Integer> st = new Stack<Integer>();
 9         for (int i=len-1; i>=0; i--) {
10             if (strs[i].length() == 0) return -1; // cases that input has two spaces, wrong input
11             if (strs[i].equals("*")) {
12                 st.push(0);
13             }
14             else {
15                 if (st.size() == 0) return -1;
16                 int num1 = st.pop();
17                 if (st.size() == 0) return -1;
18                 int num2 = st.pop();
19                 st.push(Math.max(num1, num2) + 1);
20             }
21         }
22         if (st.size() != 1) return -1;
23         return st.peek();
24     }
25 
26     public static void main (String[] args) {
27         Solution5 a = new Solution5();
28         int ret = a.treeHeight("a b * * *");
29         System.out.println(ret);
30     }
第二遍做法:仿效preorder traversal的iterative做法,用Stack, 因为有空格,先split(" "), 这样把它转化为array,且解决了上面ab * *的问题,
但是做OA的时候,大部分case过了,还是有小case过不了
2    public int treeHeight(String preorder) {
 3        if (preorder==null || preorder.length()==0) return -1;
 4        preorder.trim();
 5        if (preorder.length() == 0) return -1;
 6        String[] ar = preorder.split(" ");
 7        Stack<String> st = new Stack<String>();
 8        int maxHeight = 0;
 9        int curHeight = 0;
10        String root = ar[0];
11        int i = 0;
12        while (!root.equals("*") || !st.isEmpty()) {
13            if (!root.equals("*")) { // not null
14                st.push(root);
15                curHeight++;      // this node's height
16                maxHeight = Math.max(maxHeight, curHeight); //all time heighest
17                root = ar[++i];   // next node in preorder tranversal
18            }
19            else {
20                st.pop();
21                if (ar[i-1].equals("*")) curHeight = st.size() + 1; //only a[i]=="*" && a[i-1]=="*", curheight will change
22                root = ar[++i];
23            }
24        }
25        if (i != ar.length) return -1; // if not reaching the end of the preorder array, then the tree is not valid
26        return maxHeight;
27    }


 第21行比较tricky, 解释如下,如果遇到两个星的情况如**,表示遇到叶子节点了,pop操作之后,下一个节点(假设叫x)一定是某个节点(假设叫y)的第一个右节点,而且节点y已经不在stack里了,刚被pop掉,所以当前curHeight更新为 stack.size()+1, 这个1就是加上节点y的一层

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