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  • LeetCode -- Binary Tree Level Order Traversal II

    Question:

    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]
    ]
    

    Analysis:

    问题描述:给出一棵二叉树,写出它的从底层到最上层的层次遍历。

    思路一:一个投机取巧的思路,前面做了从上层到下层的遍历,二者的不同之处仅在于返回的列表的顺序,因此先从上到下遍历,然后将外层列表倒转一下即可。。

    /**
     * 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>>();
           List<Integer> l = new ArrayList<Integer>();
           
           if(root == null)
                       return res;
           if(root.left == null && root.right == null) {
                      l.add(root.val);
               res.add(l);
               return res;
           }
           
           TreeNode start; //记录了每一层的开始
           Queue<TreeNode> node = new LinkedList<TreeNode>();
           node.offer(root);
           while(!node.isEmpty()) {
                       TreeNode cur = node.peek();
                       List<Integer> thisLevel = new ArrayList<Integer>();
                       start = null;
                       while(node.peek() != start && !node.isEmpty()) {
                           cur = node.poll();
                           if(start == null) {
                               if(cur.left != null)
                                   start = cur.left;
                               else if(cur.right != null)
                                   start = cur.right;
                           }
                           
                           thisLevel.add(cur.val);
                           if(cur.left != null)
                               node.offer(cur.left);
                           if(cur.right != null)
                               node.offer(cur.right);
                           
                       }
                       res.add(thisLevel);
           }
           List<List<Integer>> res1 = new ArrayList<List<Integer>>();
              for(int i=0; i<res.size(); i++) {
                  res1.add(res.get(res.size() - 1 - i));
              }
           return res1;
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/little-YTMM/p/4855784.html
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