• leetcode Binary Tree Level Order Traversal I II




    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]
    ]
    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
            ArrayList<ArrayList<Integer>> result=new  ArrayList<ArrayList<Integer>>();
            if(root==null){
                return result;
            }
            LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
            queue.add(root);
            while(!queue.isEmpty()){
                ArrayList<Integer> level=new ArrayList<Integer>();
                int size=queue.size();
                for(int i=0;i<size;i++){
                    TreeNode head=queue.poll();
                    level.add(head.val);
                    if(head.left!=null){
                        queue.add(head.left);
                    }
                    if(head.right!=null){
                        queue.add(head.right);
                    }
                }
                result.add(level);
            }
            return result;
    
        }
    }

    Binary Tree Level Order Traversal  II

    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]
    ]
    /**
     * Definition for binary tree
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
             ArrayList<ArrayList<Integer>> result=new  ArrayList<ArrayList<Integer>>();
            if(root==null){
                return result;
            }
            LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
            queue.add(root);
            while(!queue.isEmpty()){
                ArrayList<Integer> level=new ArrayList<Integer>();
                int size=queue.size();
                for(int i=0;i<size;i++){
                    TreeNode head=queue.poll();
                    level.add(head.val);
                    if(head.left!=null){
                        queue.add(head.left);
                    }
                    if(head.right!=null){
                        queue.add(head.right);
                    }
                }
                result.add(0,level);
            }
            return result;
        }
    }
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  • 原文地址:https://www.cnblogs.com/lilyfindjobs/p/4068330.html
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