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],

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2
3
4
5
3
/ \
9 20
/ \
15 7

return its bottom-up level order traversal as:

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

Solution

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/**
* 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;
}
};