[LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 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
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

102. Binary Tree Level Order Traversal 的变形,只是最后输出的顺序变了,这题是从最后一层开始。

 

Python:

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if root is None:
            return []

        result, current = [], [root]
        while current:
            next_level, vals = [], []
            for node in current:
                vals.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            current = next_level
            result.append(vals)

        return result[::-1] 

C++:Iteration

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > res;
        if (root == NULL) return res;

        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            vector<int> oneLevel;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                oneLevel.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.insert(res.begin(), oneLevel);
        }
        return res;
    }
};

C++:Recursion

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > res;
        levelorder(root, 0, res);
        return vector<vector<int> > (res.rbegin(), res.rend());
    }
    void levelorder(TreeNode *root, int level, vector<vector<int> > &res) {
        if (!root) return;
        if (res.size() == level) res.push_back({});
        res[level].push_back(root->val);
        if (root->left) levelorder(root->left, level + 1, res);
        if (root->right) levelorder(root->right, level + 1, res);
    }
};  

  

类似题目:

[LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

[LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图

 

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posted @ 2018-03-20 02:53  轻风舞动  阅读(410)  评论(0编辑  收藏  举报