[LeetCode] 366. Find Leaves of Binary Tree 找二叉树的叶节点

最新推荐文章于 2025-08-09 11:46:48 发布

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最新推荐文章于 2025-08-09 11:46:48 发布

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文章标签：数据结构与算法 java python

原文链接：http://www.cnblogs.com/lightwindy/p/9583763.html

本文介绍了一种从二叉树中逐步移除并收集所有叶节点的算法实现，通过递归深度优先搜索（DFS）的方式，按层级顺序返回叶节点的列表。提供了Java、Python及C++三种语言的具体实现。

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Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty.

Example:
Given binary tree

Returns [4, 5, 3], [2], [1].

Explanation:

1. Remove the leaves [4, 5, 3] from the tree

          1
         / 
        2

2. Remove the leaf [2] from the tree

3. Remove the leaf [1] from the tree

[]

Returns [4, 5, 3], [2], [1].

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

给一个二叉树，找出它的叶节点然后删除，重复此步骤，直到二叉树为空。

The key to solve this problem is converting the problem to be finding the index of the element in the result list. Then this is a typical DFS problem on trees.

Java:

public List<List<Integer>> findLeaves(TreeNode root) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    helper(result, root);
    return result;
}
 
// traverse the tree bottom-up recursively
private int helper(List<List<Integer>> list, TreeNode root){
    if(root==null)
        return -1;
 
    int left = helper(list, root.left);
    int right = helper(list, root.right);
    int curr = Math.max(left, right)+1;
 
    // the first time this code is reached is when curr==0,
    //since the tree is bottom-up processed.
    if(list.size()<=curr){
        list.add(new ArrayList<Integer>());
    }
 
    list.get(curr).add(root.val);
 
    return curr;
}

Python:

class Solution(object):
    def findLeaves(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        def findLeavesHelper(node, result):
            if not node:
                return -1
            level = 1 + max(findLeavesHelper(node.left, result), \
                            findLeavesHelper(node.right, result))
            if len(result) < level + 1:
                result.append([])
            result[level].append(node.val)
            return level

        result = []
        findLeavesHelper(root, result)
        return result

C++:

// Time:  O(n)
// Space: O(h)

/**
 * 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>> findLeaves(TreeNode* root) {
        vector<vector<int>> result;
        findLeavesHelper(root, &result);
        return result;
    }

private:
    int findLeavesHelper(TreeNode *node, vector<vector<int>> *result) {
        if (node == nullptr) {
            return -1;
        }
        const int level = 1 + max(findLeavesHelper(node->left, result),
                                  findLeavesHelper(node->right, result));
        if (result->size() < level + 1){
            result->emplace_back();
        }
        (*result)[level].emplace_back(node->val);
        return level;
    }
};

类似题目：

[LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度

[LeetCode] 310. Minimum Height Trees 最小高度树

[LeetCode] 545. Boundary of Binary Tree 二叉树的边界