[LeetCode-interview algorithm classic-Java implementation] [102-Binary Tree Level Order Traversal (Binary Tree sequence Traversal)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [102-Binary Tree Level Order Traversal (Binary Tree sequence Traversal)], leetcode -- java
[102-Binary Tree Level Order Traversal (Binary Tree sequence Traversal )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

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:

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

Theme

Given a binary tree, It outputs the nodes of each layer.

Solutions

Use two queues: one to save the current processing layer and the other to save the next processing layer. Only process each layer.

Code Implementation

Tree node class

public class TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x) {        val = x;    }}

Algorithm Implementation class

import java.util.ArrayList;import java.util.Deque;import java.util.LinkedList;import java.util.List;public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        if (root == null) {            return result;        }        Deque<TreeNode> cur = new LinkedList<>();        Deque<TreeNode> sub = new LinkedList<>();        Deque<TreeNode> exc;        TreeNode node;        cur.addLast(root);        while (!cur.isEmpty()) {            List<Integer> layout = new LinkedList<>();            while (!cur.isEmpty()) {                node = cur.removeFirst();                layout.add(node.val);                if (node.left != null) {                    sub.addLast(node.left);                }                if(node.right != null) {                    sub.addLast(node.right);                }            }            exc = cur;            cur = sub;            sub = exc;            result.add(layout);        }        return result;    }}

Evaluation Result

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Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/47354349]

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