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:

　　藉助隊列，層序遍歷，在隊列中記錄這個節點是第幾層的。在從隊列中取節點，同層的放在一個vector中，若是新的一層的節點出現了，則把上一層的vector放入到vector的vector中。這樣記錄的是從上到下遍歷的結果，把vector反轉便可。代碼以下，經過了leetcode的在線測試。

/**
 * 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;
    }
};

時間複雜度，O(n)，n是樹的節點數。