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LeetCode: Binary Tree Level Order Traversal II [107]

2024-07-07 06:20:12   227人阅读

【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;
    }
};


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