leetcode-107-Binary-Tree-Level-Order-Traversal-II

发表于 2014-03-11 |

描述

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},

return its bottom-up level order traversal as:

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

分析

树的层次遍历，与(E) Binary Tree Level Order Traversal代码基本一样，但最后返回的是vector>(result.rbegin(), result.rend())，因为传入的是rbegin和rend，可以将序列翻转：
rbeginrend

解决方案1(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]]
        """
        if not root:
            return []
        result = []
        next_level = [root]
        
        while next_level:
            now_level = next_level
            result.insert(0, map(lambda x: x.val, now_level))
            
            next_level = []
            for item in now_level:
                if item.left:
                    next_level.append(item.left)
                if item.right:
                    next_level.append(item.right)
        return result

解决方案2(C++)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> result;
    
    void leveltravel(TreeNode* root, int level) {
        if(root == NULL) {
            return;
        }
        if(level == result.size()) {
            vector<int> v;
            result.push_back(v);
        }
        result[level].push_back(root->val);
        leveltravel(root->left, level+1);
        leveltravel(root->right, level+1);
    }

    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        leveltravel(root, 0);
        return vector<vector<int>>(result.rbegin(), result.rend());
    }
};

解决方案3(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>> result = new LinkedList<List<Integer>>();
        DFS(root, 0, result);
        return result;
    }
    
    private void DFS(TreeNode node, int i, List<List<Integer>> result) {
        if (node == null) {
            return;
        }
        
        if (i == result.size()) {
            result.add(0, new LinkedList<Integer>());
        }
        result.get(result.size()-i-1).add(node.val);
        DFS(node.left, i+1, result);
        DFS(node.right, i+1, result);
    }
}

解决方案4 （Golang）

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrderBottom(root *TreeNode) [][]int {
    if root == nil {
        return nil
    }
    result := [][]int{}
    queue := []*TreeNode{root}
    for len(queue) > 0 {
        queueLength := len(queue)
        layer := []int{}
        for i := 0; i < queueLength; i++ {
            layer = append(layer, queue[i].Val)
            if queue[i].Left != nil {
                queue = append(queue, queue[i].Left)
            }
            if queue[i].Right != nil {
                queue = append(queue, queue[i].Right)
            }
        }
        result = append(result, layer)
        queue = queue[queueLength:]
    }
    for i, resultLength := 0, len(result); i < resultLength/2; i++ {
        result[i], result[resultLength-i-1] = result[resultLength-i-1], result[i]
    }
    return result
}

相关问题

(E) Binary Tree Level Order Traversal

题目来源