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107. Binary Tree Level Order Traversal II

题目

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

解析

• 题目虽然思路很清晰，但是查看了方法2,3感觉也吸收了很多思路！

// Binary Tree Level Order Traversal II
class Solution_107 {
public:
    //bfs
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        
        vector<vector<int> > vecs;
        stack<vector<int>> sta;
        if (!root)
        {
            return vecs;
        }

        queue<TreeNode*> que;
        que.push(root);

        while (!que.empty())
        {
            int size = que.size();
            vector<int> vec;
            TreeNode* temp;
            for (int i = 0; i < size;i++)
            {
                temp = que.front();
                que.pop();

                vec.push_back(temp->val);

                if (temp->left)
                {
                    que.push(temp->left);
                }
                if (temp->right)
                {
                    que.push(temp->right);
                }
            }
            sta.push(vec);
        }

        // reverse(res.begin(),res.end());
        while (!sta.empty())
        {
            vector<int> vec = sta.top();
            sta.pop();
            vecs.push_back(vec);
        }

        return vecs;
    }

    // 记录每层的个数；curCount当前层个数，nextCount下一层个节点个数
    vector<vector<int>> levelOrderBottom1(TreeNode* root) {
        int curCount = 0, nextCount = 0, level = 0;
        vector<vector<int>> ret;
        if (!root) return ret;

        curCount = 1;
        queue<TreeNode*> q;
        q.push(root);
        ret.push_back(vector<int>(0, curCount));

        while (!q.empty()) {
            if (curCount == 0) {
                curCount = nextCount;
                nextCount = 0;
                level++;
                ret.push_back(vector<int>(0, curCount)); //初始化下一层vector的大小
            }

            TreeNode *node = q.front();
            q.pop();
            curCount--;
            ret[level].push_back(node->val); //在同一层的节点加入

            if (node->left) {
                nextCount++;
                q.push(node->left);
            }
            if (node->right) {
                nextCount++;
                q.push(node->right);
            }
        }
        reverse(ret.begin(), ret.end());
        return ret;
    }


    //dfs
    int getHeight(TreeNode *root)
    {
        if (!root) 
            return 0;
        return max(getHeight(root->left), getHeight(root->right)) + 1;
    }
    vector<vector<int> > levelOrderBottom2(TreeNode *root)
    {
        if (!root) 
            return vector<vector<int>>();

        vector<vector<int>> res(getHeight(root), vector<int>());

        dfs(root, res.size() - 1, res);
        return res;
    }
    void dfs(TreeNode *root, int height, vector<vector<int>> &res)
    {
        if (!root)
            return;
        res[height].push_back(root->val);

        dfs(root->left, height - 1, res);
        dfs(root->right, height - 1, res);
    }
};

题目来源

• 107. Binary Tree Level Order Traversal II