Massive Algorithms: LintCode 126

LintCode 126 - Max Tree

Related: LeetCode 654 - Maximum Binary Tree
Lintcode-Max Tree - LiBlog - 博客园
Given an integer array with no duplicates. A max tree building on this array is defined as follow:

The root is the maximum number in the array
The left subtree and right subtree are the max trees of the subarray divided by the root number.

Construct the max tree by the given array.

Given [2, 5, 6, 0, 3, 1], the max tree constructed by this array is:

6

/ \

  5   3

 /   / \

2   0   1

还是用到了ordered stack。对于每个数，他的左右右节点是在他与第一个比他大的数之间，比他小的最大数，因此维护一个递减stack就可以得到。

下降stack，因为遇到上升的就自动组成一个左subtree。这个stack的思想要体会。
https://segmentfault.com/a/1190000007471356

    public TreeNode maxTree(int[] A) {
        if (A == null || A.length == 0) return null;
        Stack<TreeNode> stack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            //遍历A的每个元素，创造结点node
            TreeNode node = new TreeNode(A[i]);
            //将stack中小于当前结点的结点都pop出来，存为当前结点的左子树
            while (!stack.isEmpty() && node.val >= stack.peek().val) node.left = stack.pop();
            //如果stack仍非空，剩下的结点一定大于当前结点，所以将当前结点存为stack中结点的右子树；而stack中结点本来的右子树之前已经存为当前结点的左子树了
            if (!stack.isEmpty()) stack.peek().right = node;
            //stack中存放结点的顺序为：底部为完整的max tree，从下向上是下一层孩子结点的备份，顶部是当前结点的备份
            stack.push(node);
        }
        TreeNode root = stack.pop();
        while (!stack.isEmpty()) root = stack.pop();
        return root;
    }

http://blog.welkinlan.com/2015/06/29/max-tree-lintcode-java/

https://github.com/shawnfan/LintCode/blob/master/Java/Max%20Tree.java

    public TreeNode maxTree(int[] A) {

        // write your code here

        if (A == null || A.length == 0) {

            return null;

}

        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();

        for (int i = 0; i < A.length; i++) {

            TreeNode cur = new TreeNode(A[i]);

            while (!stack.isEmpty() && cur.val > stack.peek().val) {

                cur.left = stack.pop();

}

            if (!stack.isEmpty()) {

                stack.peek().right = cur;

}

            stack.push(cur);

}

        TreeNode result = new TreeNode(0);

        while (!stack.isEmpty()) {

            result = stack.pop();

}

        return result;

}

其实这种树叫做笛卡树（ Cartesian tree）。直接递归建树的话复杂度最差会退化到O(n^2)。经典建树方法，用到的是单调堆栈。我们堆栈里存放的树，只有左子树，没有有子树，且根节点最大。
（1）如果新来一个数，比堆栈顶的树根的数小，则把这个数作为一个单独的节点压入堆栈。
（2）否则，不断从堆栈里弹出树，新弹出的树以旧弹出的树为右子树，连接起来，直到目前堆栈顶的树根的数大于新来的数。然后，弹出的那些数，已经形成了一个新的树，这个树作为新节点的左子树，把这个新树压入堆栈。

这样的堆栈是单调的，越靠近堆栈顶的数越小。
最后还要按照（2）的方法，把所有树弹出来，每个旧树作为新树的右子树

17     public TreeNode maxTree(int[] A) {
18         if (A.length==0) return null;
19 
20         Stack<TreeNode> nodeStack = new Stack<TreeNode>();
21         nodeStack.push(new TreeNode(A[0]));
22         for (int i=1;i<A.length;i++)
23             if (A[i]<=nodeStack.peek().val){
24                 TreeNode node = new TreeNode(A[i]);
25                 nodeStack.push(node);
26             } else {
27                 TreeNode n1 = nodeStack.pop();
28                 while (!nodeStack.isEmpty() && nodeStack.peek().val < A[i]){
29                     TreeNode n2 = nodeStack.pop();
30                     n2.right = n1;
31                     n1 = n2;
32                 }
33                 TreeNode node = new TreeNode(A[i]);
34                 node.left = n1;
35                 nodeStack.push(node);
36             }
37         
38 
39         TreeNode root = nodeStack.pop();
40         while (!nodeStack.isEmpty()){
41             nodeStack.peek().right = root;
42             root = nodeStack.pop();
43         }
44 
45         return root;
51     }

https://codesolutiony.wordpress.com/2015/01/28/lintcode-max-tree/

    public TreeNode maxTree(int[] A) {

        Stack<TreeNode> stack = new Stack<TreeNode>();//decreasing stack

        for (int i = 0; i < A.length; i++) {

            TreeNode newNode = new TreeNode(A[i]);

            TreeNode pre = null;

            while (!stack.isEmpty() && stack.peek().val < A[i]) {

                TreeNode cur = stack.pop();

                if (pre != null) {

                    cur.right = pre;

}

                pre = cur;

                newNode.left = pre;

}

            stack.push(newNode);

}

        TreeNode preNode = null;

        while (!stack.isEmpty()) {

            TreeNode curNode = stack.pop();

            curNode.right = preNode;

            preNode = curNode;

}

        return preNode;

}

https://www.jianshu.com/p/e05e598c8073

通过观察发现规律，对于每个node的父亲节点 = min(左边第一个比它大的，右边第一个比它大的)
维护一个降序数组，可以实现对这个min的快速查找

# stack = [2], push 5 因为 5 > 2, 所以2是5的左儿子, pop 2
# stack = [], push 5
# stack = [5], push 6, 因为 6 > 5，所以5是6的左儿子, pop 5
# stack = [], push 6
# stack = [6], push 0, 因为0 < 6, stack = [6], 所以0是6的右儿子
# stack = [6, 0], push 3, 因为3 > 0, 所以0是3的左儿子， pop 0
# stack = [6], 所以3是6的右儿子， push 3
# stack = [6, 3], push 1, 因为1 < 3，所以1是3的右儿子

    def maxTree(self, A):
        # write your code here
        stack = []
        for element in A:
            node = TreeNode(element)
            while len(stack) != 0 and element > stack[-1].val:
                node.left = stack.pop()
            if len(stack) != 0:
                stack[-1].right = node
            stack.append(node)
        return stack[0]

X. Recursive
https://segmentfault.com/a/1190000007471356

    public TreeNode maxTree(int[] A) {
        if (A == null || A.length == 0) return null;
        return buildMax(A, 0, A.length-1);
    }
    public TreeNode buildMax(int[] A, int start, int end) {
        if (start > end) return null;
        int max = Integer.MIN_VALUE;
        int maxIndex = -1;
        for (int i = start; i <= end; i++) {
            if (A[i] >= max) {
                max = A[i];
                maxIndex = i;
            }
        }
        TreeNode root = new TreeNode(max);
        root.left = buildMax(A, start, maxIndex-1);
        root.right = buildMax(A, maxIndex+1, end);
        return root;
    }

https://www.cnblogs.com/lishiblog/p/4187936.html

17     public TreeNode maxTree(int[] A) {
18         if (A.length==0) return null;
19 
20         TreeNode root = new TreeNode(A[0]);
21         for (int i=1;i<A.length;i++)
22             if (A[i]>root.val){
23                 TreeNode node = new TreeNode(A[i]);
24                 node.left = root;
25                 root = node;
26             } else insertNode(root,null,A[i]);
27 
28         return root;
29     }
30 
31     public void insertNode(TreeNode cur, TreeNode pre, int val){
32         if (cur==null){
33             TreeNode node = new TreeNode(val);
34             pre.right = node;
35             return;
36         }
37 
38         if (cur.val<val){
39             TreeNode node = new TreeNode(val);
40             pre.right = node;
41             node.left = cur;
42             return;
43         } else 
44             insertNode(cur.right,cur,val);
45     }

https://richdalgo.wordpress.com/2015/01/28/lintcode-max-tree/
http://dinnerwanzi.blogspot.com/2015/07/lintcode-94-binary-tree-max-sum.html

    int maxPathSum(TreeNode *root) {
        // write your code here
        if (!root)
            return 0;
        myMax=INT_MIN;
        maxHelper(root);
        return myMax;
    }
    
    int maxHelper(TreeNode* root){
        if (!root)
            return 0;
        int l=max(0, maxHelper(root->left));
        int r=max(0, maxHelper(root->right));
        myMax=max(myMax, l+r+root->val);
        return max(l,r)+root->val;
    }
    int myMax;

Read full article from Lintcode-Max Tree - LiBlog - 博客园

LintCode 126 - Max Tree

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