題目鏈接 : https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/

題目描述:

給定一個二叉樹，返回其節(jié)點值自底向上的層次遍歷。 （即按從葉子節(jié)點所在層到根節(jié)點所在的層，逐層從左向右遍歷）

例如：

給定二叉樹 [3,9,20,null,null,15,7],

    3

   / \

  9  20

    /  \

   15   7

返回其自底向上的層次遍歷為：

[

  [15,7],

  [9,20],

  [3]

]

思路:

與上一題層次遍歷一樣,只不過輸出的順序取反了!

所以只需要從頭添加數(shù)組就可以了!

思路一: 迭代

思路二: 遞歸

代碼:

思路一:

# Definition for a binary tree node.

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:

        from collections import deque

        if not root: return []

        queue = deque()

        queue.appendleft(root)

        res = []

        while queue:

            tmp = []

            n = len(queue)

            for _ in range(n):

                node = queue.pop()

                tmp.append(node.val)

                if node.left:

                    queue.appendleft(node.left)

                if node.right:

                    queue.appendleft(node.right)

            res.insert(0, tmp)

        return res

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>> res = new LinkedList<>();

        if (root == null) return res;

        Deque<TreeNode> queue = new LinkedList<>();

        queue.add(root);

        while (!queue.isEmpty()) {

            List<Integer> tmp = new ArrayList<>();

            int n = queue.size();

            for (int i = 0; i < n; i++) {

                TreeNode node = queue.poll();

                tmp.add(node.val);

                if (node.left != null) queue.add(node.left);

                if (node.right != null) queue.add(node.right);

            }

            res.add(0, tmp);

        }

        return res;

    }

}

思路二:

# Definition for a binary tree node.

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:

        res = []

        def helper(root, depth):

            if not root: return

            if depth == len(res):

                res.insert(0, [])

            res[-(depth+1)].append(root.val)

            helper(root.left, depth+1)

            helper(root.right, depth+1)

        helper(root, 0)

        return res

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>> res = new LinkedList<>();

        helper(res, root, 0);

        return res;

    }

    private void helper(List<List<Integer>> res, TreeNode root, int depth) {

        if (root == null) return;

        if (res.size() == depth) res.add(0, new ArrayList<>());

        res.get(res.size() - depth - 1).add(root.val);

        helper(res, root.left, depth + 1);

        helper(res, root.right, depth + 1);

    }

}

總結(jié)

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