• 0107. Binary Tree Level Order Traversal II (E)


    Binary Tree Level Order Traversal II (E)

    题目

    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,null,null,15,7],

        3
       / 
      9  20
        /  
       15   7
    

    return its bottom-up level order traversal as:

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

    题意

    按从下到上的顺序记录二叉树的每一层。

    思路

    BFS后再将结果集逆序。


    代码实现

    Java

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<List<Integer>> levelOrderBottom(TreeNode root) {
            List<List<Integer>> ans = new ArrayList<>();
            Queue<TreeNode> q = new ArrayDeque<>();
    
            if (root != null) {
                q.offer(root);
            }
    
            while (!q.isEmpty()) {
                List<Integer> list = new ArrayList<>();
                int size = q.size();
                for (int i = 0; i < size; i++) {
                    TreeNode cur = q.poll();
                    list.add(cur.val);
                    if (cur.left!=null) q.offer(cur.left);
                    if (cur.right!=null) q.offer(cur.right);
                }
                ans.add(list);
            }
            
            Collections.reverse(ans);
            return ans;
        }
    }
    

    JavaScript

    /**
     * Definition for a binary tree node.
     * function TreeNode(val, left, right) {
     *     this.val = (val===undefined ? 0 : val)
     *     this.left = (left===undefined ? null : left)
     *     this.right = (right===undefined ? null : right)
     * }
     */
    /**
     * @param {TreeNode} root
     * @return {number[][]}
     */
    var levelOrderBottom = function (root) {
      let bfs = []
      let q = []
      if (root) {
        q.push(root)
      }
      while (q.length > 0) {
        let size = q.length
        let tmp = []
        while (size > 0) {
          let p = q.shift()
          tmp.push(p.val)
          if (p.left) q.push(p.left)
          if (p.right) q.push(p.right)
          size--
        }
        bfs.push(tmp)
      }
      return bfs.reverse()
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/13228051.html
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