107. Binary Tree Level Order Traversal II

2024年1月3日 193次阅读

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]
]

 

方法1：分层考虑，直接用vector

/**
 * 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) {
        //每层分开考虑
        vector<vector<int>> res;
        addValue(0, res,root);
        reverse(res.begin(),res.end());
        return res;
    }
    void addValue(int count, vector<vector<int>> &res,TreeNode* root){
        if(root==NULL) return;
        if(res.size()<count+1) res.push_back({root->val});
        else res[count].push_back(root->val);
        addValue(count+1,res,root->left);
        addValue(count+1,res,root->right);
    }
    /*单独节点分开考虑的
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        map<int,vector<int>> treeValue;
        vector<vector<int>> res;
        addValue(0,treeValue,root);
        for(auto a:treeValue){
            res.insert(res.begin(),a.second);
        }
        return res;
    }
    
    void addValue(int count, map<int,vector<int>> &treeValue,TreeNode* root){
        if(root==NULL) return;
        if(treeValue.count(count)) treeValue.find(count)->second.push_back(root->val);
        else treeValue[count]={root->val};
        addValue(count+1,treeValue,root->left);
        addValue(count+1,treeValue,root->right);
    }*/
};

Tips：递归传值时，是count+1，不能用count++,这个会改变当前count的值。

 采用了reverse()函数

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