[LeetCode] 107. binary tree hierarchy traversal II

Topic links: https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/

Subject description:

Given a binary tree, the node returns its bottom-up value hierarchy traversal. (Ie, by physical layer to the layer from the leaf nodes of the root node, layer by layer traversal from left to right)

For example:
given binary tree [3,9,20, null, null, 15,7 ],

    3
   / \
  9  20
    /  \
   15   7

Returns to its bottom-up hierarchy traversal is:

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

Ideas:

And the previous question traverse the level , as the order but the output negated!

You only need to add an array from scratch on it!

A thought: Iteration

Ideas II: recursive

Code:

A thought:

# 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;  
    }
}

Thinking two:

# 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);
    }
}