Binary Tree Level Order Traversal II

问题描述:

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).

For example:
Given binary tree {3,9,20,#,#,15,7},

  3
 / \
9  20
  /  \
 15   7

return its bottom-up level order traversal as:

1
2
3
4
5
[
[15,7],
[9,20],
[3]
]

confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

1
2
3
4
5
6
7
  1
/ \
2 3
/
4
\
5

The above binary tree is serialized as “{1,2,3,#,#,4,#,#,5}”.

解决思路:

和前面一样

代码:

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/**
* Definition for binary tree
* 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>> LOB;
if (root == NULL)
return LOB;
int level = 0;
int count = 1;
queue<TreeNode *> node;
node.push(root);
stack<vector<int>> stk;
vector<int> sub(0);
while(!node.empty()){
sub.clear();
level = 0;
for(int i = 0; i < count; i++){
root = node.front();
node.pop();
sub.push_back(root->val);
if(root->left != NULL){
node.push(root->left);
++level;
}
if(root->right != NULL){
node.push(root->right);
++level;
}
}
stk.push(sub);
count = level;
}
while(!stk.empty()){
vector<int> tmp = stk.top();
LOB.push_back(tmp);
stk.pop();
}
return LOB;
}
};