Binary Tree Level Order Traversal

Question:

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]

]

Solution:

　　借助队列，层序遍历，在队列中记录这个节点是第几层的。在从队列中取节点，同层的放在一个vector中，如果新的一层的节点出现了，则把上一层的vector放入到vector的vector中。这样记录的是从上到下遍历的结果，把vector反转即可。代码如下，通过了leetcode的在线测试。

/**

 * 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> > vecRes;

        if (!root)

            return vecRes;

            

        queue<pair<TreeNode*, int> > nodeQue; // first is val, second is layer

        vector<vector<int> > vecWorker;

        vector<int> vec;

        nodeQue.push(make_pair(root, 1));

        int layer = 1;

        

        while (!nodeQue.empty()) {

            TreeNode* node = nodeQue.front().first;

            int nodeLayer = nodeQue.front().second;

            nodeQue.pop();

            if (nodeLayer > layer) {

                vecWorker.push_back(vec);

                vec.clear();

                layer = nodeLayer;

            }

            vec.push_back(node->val);

            if (node->left) {

                nodeQue.push(make_pair(node->left, nodeLayer+1));

            }

            if (node->right) {

                nodeQue.push(make_pair(node->right, nodeLayer+1));

            }

        }

        vecWorker.push_back(vec);

        

        for (int i = vecWorker.size()-1; i >= 0; i--) {

            vecRes.push_back(vecWorker[i]);

        }

        

        return vecRes;

    }

};

时间复杂度，O(n)，n是树的节点数。