LeetCode:Binary Tree Level Order Traversal

3
/ 
9
20
/

15
7

[
[3],
[9,20],
[15,7]
]

思路：

层次遍历，BFS。

新

java code:

/**
* Definition for a binary tree node.
* public class TreeNode {
*
int val;
*
TreeNode left;
*
TreeNode right;
*
TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<TreeNode>();
List<List<Integer>> ans = new LinkedList<List<Integer>>();
if(root == null) return ans;
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
List<Integer> subAns = new LinkedList<Integer>();
for(int i=0;i<size;i++) {
TreeNode tmp = queue.poll();
subAns.add(tmp.val);
if(tmp.left != null) queue.offer(tmp.left);
if(tmp.right != null) queue.offer(tmp.right);
}
ans.add(subAns);
}
return ans;
}
}

c++ code:

/**
* 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>> levelOrder(TreeNode* root) {
vector<vector<int>> ret;
if(NULL == root) return ret;
queue<TreeNode *> q[2];
stack<vector<int>> s;
int cur=0;
q[cur].push(root);
while(!q[cur].empty()) {
vector<int> tmp;
while(!q[cur].empty()) {
TreeNode *p = q[cur].front();
tmp.push_back(p->val);
if(p->left) q[cur^1].push(p->left);
if(p->right) q[cur^1].push(p->right);
q[cur].pop();
}
cur^=1;
ret.push_back(tmp);
}
return ret;
}
};

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