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  • LeetCode OJ--Binary Tree Level Order Traversal II

    http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

    树的层序遍历,和上一道题相比,对结果做一个顺序调整 reverse()

    /**
     * Definition for binary tree
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
            vector<vector<int> > levelOrderBottom(TreeNode *root) {
            vector<vector<int> > ans;
            if(root == NULL)
                return ans;
            int num = 1,num2 = 1,nullNum = 0,nullNumAcc = 0;
            queue<TreeNode *> myQueue;
            myQueue.push(root);
            TreeNode *nodeFront;
    
            vector<int> onePiece;
            while(!myQueue.empty())
            {
                nodeFront = myQueue.front();
                myQueue.pop();
                num--;
                 
                onePiece.push_back(nodeFront->val);
                if(nodeFront->left)
                    myQueue.push(nodeFront->left);
                else
                    nullNum++;
                if(nodeFront->right)
                    myQueue.push(nodeFront->right);
                else
                    nullNum++;
                 
                if(num == 0)
                {
                    if(onePiece.empty())
                        break;
                    ans.push_back(onePiece);
                    onePiece.clear();
                    num2 = num2*2;
                    nullNumAcc = nullNumAcc*2 + nullNum;
                    num = num2 - nullNumAcc; 
                    nullNum = 0;
                }
            }
            reverse(ans.begin(),ans.end());
            return ans; 
        }        
    };
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  • 原文地址:https://www.cnblogs.com/qingcheng/p/3550449.html
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