原题

LintCode 70. 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]
]

代码

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
    * @param root: A tree
    * @return: buttom-up level order a list of lists of integer
    */
    vector<vector<int>> levelOrderBottom(TreeNode * root) {
        // write your code here
        vector<vector<int>> res;
        if (root == NULL) return res;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            int size = q.size();
            vector<int> level;
            for (int i = 0; i < size; i++) {
                TreeNode *cur = q.front();
                q.pop();
                level.push_back(cur->val);
                if (cur->left) q.push(cur->left);
                if (cur->right) q.push(cur->right);
            }
            res.insert(res.begin(), level);
        }
        return res;
    }
};