LeetCode 0107 - Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

Desicription

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],

代码语言：javascript复制

  3
 / 
9  20
  /  
 15   7

return its bottom-up level order traversal as:

代码语言：javascript复制

[
  [15,7],
  [9,20],
  [3]
]

Solution

代码语言：javascript复制

/**
 * 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 {
private:
    vector<vector<int>> res;
    void searchTree(TreeNode* root, int level) {
        if(!root)
            return ;
        if(level == res.size())
            res.push_back(vector<int>());
        res[level].push_back(root->val);
        searchTree(root->left, level 1);
        searchTree(root->right, level 1);
    }
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        searchTree(root, 0);
        for(int i = 0, j = res.size()-1; i <= j; i  , j--)
            swap(res[i], res[j]);
        return res;
    }
};

binary null return traversal tree

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