366. 寻找二叉树的叶子节点 🔒

题目描述

给你一棵二叉树的 root 节点，请按照以下方式收集树的节点：

收集所有的叶子节点。
移除所有的叶子节点。
重复以上步骤，直到树为空。

示例 1：

输入：root = [1,2,3,4,5]
输出：[[4,5,3],[2],[1]]
解释：
[[3,5,4],[2],[1]] 和 [[3,4,5],[2],[1]] 也被视作正确答案，因为每一层返回元素的顺序不影响结果。

示例 2：

输入：root = [1]
输出：[[1]]

提示：

树中节点的数量在[1, 100]范围内。
-100 <= Node.val <= 100

解法

方法一：DFS

我们可以使用深度优先搜索的方法，递归遍历二叉树，将每个节点的高度作为索引，将节点的值添加到对应索引的数组中。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

Python3JavaC++GoTypeScriptC#

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findLeaves(self, root: Optional[TreeNode]) -> List[List[int]]:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            h = max(l, r)
            if len(ans) == h:
                ans.append([])
            ans[h].append(root.val)
            return h + 1

        ans = []
        dfs(root)
        return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<List<Integer>> ans = new ArrayList<>();

    public List<List<Integer>> findLeaves(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        int h = Math.max(l, r);
        if (ans.size() == h) {
            ans.add(new ArrayList<>());
        }
        ans.get(h).add(root.val);
        return h + 1;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> findLeaves(TreeNode* root) {
        vector<vector<int>> ans;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return 0;
            }
            int l = dfs(root->left);
            int r = dfs(root->right);
            int h = max(l, r);
            if (ans.size() == h) {
                ans.push_back({});
            }
            ans[h].push_back(root->val);
            return h + 1;
        };
        dfs(root);
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findLeaves(root *TreeNode) (ans [][]int) {
    var dfs func(*TreeNode) int
    dfs = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        l, r := dfs(root.Left), dfs(root.Right)
        h := max(l, r)
        if len(ans) == h {
            ans = append(ans, []int{})
        }
        ans[h] = append(ans[h], root.Val)
        return h + 1
    }
    dfs(root)
    return
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function findLeaves(root: TreeNode | null): number[][] {
    const ans: number[][] = [];
    const dfs = (root: TreeNode | null): number => {
        if (root === null) {
            return 0;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        const h = Math.max(l, r);
        if (ans.length === h) {
            ans.push([]);
        }
        ans[h].push(root.val);
        return h + 1;
    };
    dfs(root);
    return ans;
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList<IList<int>> FindLeaves(TreeNode root) {
        var ans = new List<IList<int>>();

        int Dfs(TreeNode node) {
            if (node == null) {
                return 0;
            }
            int l = Dfs(node.left);
            int r = Dfs(node.right);
            int h = Math.Max(l, r);
            if (ans.Count == h) {
                ans.Add(new List<int>());
            }
            ans[h].Add(node.val);
            return h + 1;
        }

        Dfs(root);
        return ans;
    }
}

366. 寻找二叉树的叶子节点 🔒

题目描述

解法

方法一：DFS

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