107. Binary Tree Level Order Traversal II

難度:Medium

Python:

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
def dfs(root, depth, result):
    if root == None:
        return
    if depth == len(result):
        result.append([])
        
    result[depth].append(root.val)
    
    dfs(root.left, depth+1, result)
    dfs(root.right, depth+1, result)
    
    
class Solution:
    def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        result = []
        dfs(root,0,result)
        result.reverse()
        return result

C++:

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void dfs(TreeNode* root, int depth, vector<vector<int>>& result) {
    if(root == nullptr){
        return;
    }
    if(depth == result.size()) {
        result.push_back({});
    }
    
    result[depth].push_back(root->val);
    
    dfs(root->left,depth+1,result);
    dfs(root->right, depth+1,result);
}
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        dfs(root, 0, result);
        reverse(begin(result), end(result));
        return result;
    }
};