Binary Tree Level Order Traversal II 题解

题目来源：Binary Tree Level Order Traversal II

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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,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

解题思路：

跟前一题Binary Tree Level Order Traversal唯一的区别就是这个将最后结果reverse一下。 这里就只列了其中一种代码了。

常规方法, 两个queue交替

参见Binary Tree Level Order Traversal。

单queue+隔板

前面word ladder ii就提到过bfs，用隔板将各层之间隔离出来。只用一个queue就能知道某层是否已经遍历完毕。

     vector<vector<int> > levelOrderBottom(TreeNode *root) 
    {
        vector<vector<int> > result;
        if(root == NULL) return result;
        queue<TreeNode*> q;
        q.push(root);
        while(! q.empty())
        {
            q.push(NULL);
            vector<int> level;
            while(q.front() != NULL)
            {
                auto node = q.front(); q.pop();
                level.push_back(node->val);
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
            result.push_back(level);
            q.pop();//pop NULL
        }
        std::reverse(result.begin(), result.end());
        return move(result);
    }

递归

参见Binary Tree Level Order Traversal。