LeetCode: Binary Tree Level Order Traversal II [107]

【题目】

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]
]

confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.

【题意】

逐层遍历二叉树,每层的值保存到一个vector。与Binary Tree Level Order Traversal不同的地方在于,本题要求从二叉树的底层开始向上输出。

【思路】

先从上到下逐层输出,保存到栈中。在恢复成从向往上的顺序输出。

【代码】

/**
 * 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> >result;
        if(root==NULL)return result;

        stack<vector<int>> st_result;
        queue<TreeNode*>q1;
        queue<TreeNode*>q2;
        //初始化q1
        q1.push(root);
        while(!q1.empty() || !q2.empty()){
            vector<int>sequence;
            if(!q1.empty()){
                while(!q1.empty()){
                    TreeNode*node = q1.front(); q1.pop();
                    sequence.push_back(node->val);
                    //将下层节点保存到q2
                    if(node->left)q2.push(node->left);
                    if(node->right)q2.push(node->right);
                }
            }
            else{
                while(!q2.empty()){
                    TreeNode*node = q2.front(); q2.pop();
                    sequence.push_back(node->val);
                    //将下层节点保存到q2
                    if(node->left)q1.push(node->left);
                    if(node->right)q1.push(node->right);
                }
            }
            st_result.push(sequence);
        }
        //调转输出顺序
        while(!st_result.empty()){
            result.push_back(st_result.top());
            st_result.pop();
        }
        return result;
    }
};

LeetCode: Binary Tree Level Order Traversal II [107]

时间: 2024-08-05 11:39:17

LeetCode: Binary Tree Level Order Traversal II [107]的相关文章

[leetcode]Binary Tree Level Order Traversal II @ Python

原题地址:http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/ 题意: 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

LeetCode——Binary Tree Level Order Traversal II

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 trave

[LeetCode]Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II 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 re

LeetCode:Binary Tree Level Order Traversal II (按层遍历)

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 trave

LeetCode &quot;Binary Tree Level Order Traversal II&quot; using DFS

BFS solution is intuitive - here I will show a DFS based solution: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solut

LeetCode Binary Tree Level Order Traversal II (二叉树颠倒层序)

题意:从左到右统计将同一层的值放在同一个容器vector中,要求上下颠倒,左右不颠倒. 思路:广搜逐层添加进来,最后再反转. 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9

leetCode 107. Binary Tree Level Order Traversal II 二叉树层次遍历反转

107. Binary Tree Level Order Traversal II 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  

【Leetcode】Binary Tree Level Order Traversal II

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 trave

Leetcode 树 Binary Tree Level Order Traversal II

本文为senlie原创,转载请保留此地址:http://blog.csdn.net/zhengsenlie Binary Tree Level Order Traversal II Total Accepted: 10080 Total Submissions: 32610 Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, l