Binary Tree Level Order Traversal II

Description:

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).

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]
]

分析：

思路：

Code:

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/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 
 
class Solution {
    /**
     * @param root : The root of binary tree.
     * @return : buttom-up level order a list of lists of integer
     */
public:
    vector<vector<int>> levelOrderBottom(TreeNode *root) {
        // write your code here
        if (root == nullptr) return vector<vector<int> >();
        
        vector<pair<TreeNode*,int> > q;
        int count = 0;
        int level = 1;
        q.push_back(make_pair(root, level));
        
        while (count < q.size()) {
            TreeNode *p = q[count].first;
            level = q[count].second;
            
            if (p->left) q.push_back(make_pair(p->left, level + 1));
            if (p->right) q.push_back(make_pair(p->right, level + 1));
            count++;
        }
        
        vector<vector<int> > rs(level, vector<int>());
        for (int i = 0; i < q.size(); i++) {
            rs[level-q[i].second].push_back(q[i].first->val);
        }
        
        return rs;
    }
};