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] ]
两种思路 dfs和bfs都可以
dfs
/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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/*vector<vector<int>> levelOrderBottom(TreeNode* root) {
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vector<vector<int>>res;
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dfs(root,0,res);
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reverse(res.begin(), res.end());
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return res;
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}
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vector<vector< int>>& dfs(TreeNode*root,int level,vector<vector<int>> &res)
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{
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if(root=NULL)
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return res;
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if(level==res.size())//res存储的上一次所包含的，如果有左右结点，那么下一次的level必定要使res新加一层
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{
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res.push_back(vector<int>());
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}
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res[level].push_back(root->val);
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dfs(root->left,level+1,res);
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dfs(root->right,level+1,res);
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}*/
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vector<vector<int> > res;
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void DFS(TreeNode* root, int level)
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{
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if (root == NULL) return;
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if (level == res.size()) // The level does not exist in output
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{
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res.push_back(vector<int>()); // Create a new level
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}
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res[level].push_back(root->val); // Add the current value to its level
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DFS(root->left, level+1); // Go to the next level
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DFS(root->right,level+1);
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}
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vector<vector<int> > levelOrderBottom(TreeNode *root) {
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DFS(root, 0);
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return vector<vector<int> > (res.rbegin(), res.rend());
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}
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};
bfs
DFS solution:
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
if(root == null) return wrapList;
queue.offer(root);
while(!queue.isEmpty()){
int levelNum = queue.size();
List<Integer> subList = new LinkedList<Integer>();
for(int i=0; i<levelNum; i++) {
if(queue.peek().left != null) queue.offer(queue.peek().left);
if(queue.peek().right != null) queue.offer(queue.peek().right);
subList.add(queue.poll().val);
}
wrapList.add(0, subList);
}
return wrapList;
}
}