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Leetcode[102, 107] Binary Tree Level Order Traversal

07 Nov 2015

###Task1 Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ]

###Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        ret = []
        if not root:
            return ret
        queue = collections.deque()
        queue.append(root)
        while len(queue) > 0:
            size = len(queue)
            res = []
            for i in range(size):
                cur = queue.popleft()
                res.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            ret.append(res)
        return ret

###Java

/**
 * 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) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> temp = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                temp.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(temp);
        }
        
        return res;
    }
}

###Points

###Task2 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] ] ###Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        ret = []
        if not root:
            return ret
        queue = collections.deque()
        queue.append(root)
        while len(queue) > 0:
            size = len(queue)
            res = []
            for i in range(size):
                cur = queue.popleft()
                res.append(cur.val)
                if cur.left:
                    queue.append(cur.left)
                if cur.right:
                    queue.append(cur.right)
            ret.append(res)
        return ret[::-1]

###Java

/**
 * 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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> temp = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                temp.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(temp);
        }
        Collections.reverse(res);
        return res;
    }
}

###Points

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