leetcode654 leetcode654 最大二叉树（利用vector创建）

[3,2,1,6,0,5]

stk is empty

push 3

2<3 3->(right)2 push 2

stk 3 2

1<2 2->(right)1 push 1

stk 3 2 1

---

6现在是最大的，拿6和stk里面的各个元素进行比较，将小于6的作为左孩子。stk里面的元素已经存在了树结构关系。

6>1 6->(left)1 pop 1

stk 3 2

6>2 6->(left)2 pop2

stk 3

6>3 6->(left)3 pop3

stk is empty

6->(left)3->(right)2->(right)1

push 6

-------------------------

0<6 6->(right)0 push 0

stk 6 0

重复上面的步骤，就可以得到一个最大二叉树

 

 

/**

* Definition for a binary tree node.

* struct TreeNode {

*     int val;

*     TreeNode *left;

*     TreeNode *right;

*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

* };

*/

class Solution {

public:

    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {

        if(nums.size() == 0)

            return NULL;

        vector<TreeNode*> stk;

        for(int i = 0;i<nums.size();i++){

            TreeNode* cur = new TreeNode(nums[i]);

            // stk 中存放 3 2 1，且3->2->1,现在有6,循环6->1,6->2,6->3,until stk is empty.6->3->2->1

            while(stk.size()&&(stk.back()->val < cur->val)){

                cur->left = stk.back();

                stk.pop_back();

            }

            if(stk.size()){// stk.size()&&(stk.back()->val > cur->val)

                stk.back()->right = cur;

            }

            stk.push_back(cur);

            // stk only 6

        }

        return stk.front();

    }

};