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107.Binary Tree Level Order Traversal II

时间:2017-07-22 19:49:38      阅读:122      评论:0      收藏:0      [点我收藏+]

标签:front   traversal   push   des   ever   tree node   public   str   --   

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,null,null,15,7],

    3
   /   9  20
    /     15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
}
/**
 * Definition for a binary tree node.
 * 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) {
        queue<TreeNode *> q;
        vector<vector<int> > vv;
        if (root != NULL) {
            q.push(root);
        }
        while(!q.empty()) { 
            int m = q.size();
            vector<int> v;
            while(m > 0) {
                TreeNode * n = q.front();q.pop();
                v.push_back(n->val);
                TreeNode * left = n->left;
                TreeNode * right = n->right;
                if (left != NULL) q.push(left);
                if (right != NULL) q.push(right);
                m--;
            }
            if(v.size())vv.push_back(v);
        }
        reverse(vv.begin(), vv.end());
        return vv;
    }
};

树的层次遍历,输出每一层

107.Binary Tree Level Order Traversal II

标签:front   traversal   push   des   ever   tree node   public   str   --   

原文地址:http://www.cnblogs.com/pk28/p/7222019.html

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