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:
[
[15,7],
[9,20],
[3]
]
这道题是求二叉树从下往上的层次遍历,与Binary Tree Level Order Traversal十分相似,只是在最后一步将ans反转一下就可以了。
直接贴上代码了:
/**
* 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> > ans;
queue<TreeNode*> q;
if (root == NULL)
return ans;
q.push(root);
int nextCount = 0;
int curCount = 1;
vector<int> temp;
while (!q.empty()){
TreeNode* p = q.front();
q.pop();
temp.push_back(p->val);
if (p->left){
q.push(p->left);
nextCount++;
}
if (p->right){
q.push(p->right);
nextCount++;
}
curCount--;
if (!curCount){
ans.push_back(temp);
temp=vector<int>();
curCount = nextCount;
nextCount = 0;
}
}
reverse(ans.begin(), ans.end());
return ans;
}
};
[LeetCode]Binary Tree Level Order Traversal II
原文:http://blog.csdn.net/kaitankedemao/article/details/43800765