题目地址：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/open in new window
Total Accepted: 55876 Total Submissions: 177210 Difficulty: Easy
题目描述
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).
Forexample:
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]
]
题目大意
从叶子向根进行层次遍历。
解题方法
方法一：DFS
通过直接的每行元素放到一个List中，再把List放到要返回的List中的方法。最后需要翻转。
标准的DFS解法如下。我都已经背会了。
Python代码如下：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
self.levelOrder(root, res, 0)
return res[::-1]
def levelOrder(self, root, res, level):
if not root: return
if level >= len(res):
res.append([])
res[level].append(root.val)
self.levelOrder(root.left, res, level + 1)
self.levelOrder(root.right, res, level + 1)
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下面的C++代码没有使用反转结果数组的方式，而是先算了树的高度，然后把第一层的节点放到结果数组的最后，这样实现了逆序。
/**
* 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) {
int d = depth(root);
vector<vector<int>> res(d);
dfs(root, res, d - 1);
return res;
}
void dfs(TreeNode* root, vector<vector<int>>& res, int depth) {
if (!root) return;
res[depth].push_back(root->val);
dfs(root->left, res, depth - 1);
dfs(root->right, res, depth - 1);
}
int depth(TreeNode* root) {
if (!root) return 0;
return max(depth(root->left), depth(root->right)) + 1;
}
};
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方法二：迭代
迭代方法也很经典，把每层的节点都放进一个队列里，把每一层的节点依次退出队列，把节点值放入这一层的结果中，并且在队列中放入下一层的节点。最后判断这一层的结果不是空的话，那么就放到最终结果里去。最后，把最终结果翻转。
python代码如下：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res = []
que = collections.deque()
que.append(root)
level = 0
while que:
levelVal = []
size = len(que)
for _ in range(size):
node = que.popleft()
if not node:
continue
levelVal.append(node.val)
que.append(node.left)
que.append(node.right)
level += 1
if levelVal:
res.append(levelVal)
return res[::-1]
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C++代码如下：
/**
* 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) {
vector<vector<int>> res;
queue<TreeNode*> que;
que.push(root);
while (!que.empty()) {
vector<int> level;
int size = que.size();
while (size --) {
TreeNode* node = que.front(); que.pop();
if (!node) continue;
level.push_back(node->val);
que.push(node->left);
que.push(node->right);
}
if (!level.empty()) {
res.push_back(level);
}
}
reverse(res.begin(), res.end());
return res;
}
};
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