README.md

Binary Tree Level Order Traversal II

Description

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

Tags: Tree, Breadth-first Search

思路 0

题意是从下往上按层遍历二叉树，每一层是从左到右，按层遍历，很明显，宽搜第一时间符合，因为是从下往上，所以插入的时候每次插到链表头即可。

/**
 * 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) {
        if (root == null) return Collections.emptyList();
        List<List<Integer>> list = new LinkedList<>();
        LinkedList<TreeNode> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()) {
            int size = q.size();
            List<Integer> sub = new LinkedList();
            for(int i = 0; i < size; ++i) {
                TreeNode node = q.remove();
                sub.add(node.val);
                if (node.left != null) q.add(node.left);
                if (node.right != null) q.add(node.right);
            }
            list.add(0, sub);
        }
        return list;
    }
}

思路 1

另一种思路就是深搜，深搜的时候同时记录深度，然后在相应的层插入节点值即可。

/**
 * 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>> list = new LinkedList<>();
        helper(list, root, 0);
        return list;
    }

    private void helper(List<List<Integer>> list, TreeNode root, int level) {
        if (root == null) return;
        if (level >= list.size()) {
            list.add(0, new LinkedList<>());
        }
        helper(list, root.left, level + 1);
        helper(list, root.right, level + 1);
        list.get(list.size() - level - 1).add(root.val);
    }
}

kotlin(248ms/100.00%):

class Solution {

    private fun turn(list: ArrayList<ArrayList<Int>>, node: TreeNode?, depth: Int) {
        if (node == null) return
        if (list.size <= depth) list.add(ArrayList())
        list[depth].add(node.`val`)
        turn(list, node.left, 1 + depth)
        turn(list, node.right, 1 + depth)
    }

    fun levelOrderBottom(root: TreeNode?): List<List<Int>> {
        val list = ArrayList<ArrayList<Int>>()
        turn(list, root, 0)
        list.reverse()
        return list
    }
}

JavaScript：

var levelOrderBottom = function(root) {
    if (root === null) { return []; }
    let result = [];
    let queue = [root];
    while(queue.length > 0) {
        let size = queue.length;
        let current = [];
        for (let i = 0; i < size; i++) {
            let head = queue.shift();
            current.push(head.val);
            if(head.left !== null) { queue.push(head.left); }
            if(head.right !== null) { queue.push(head.right); }
        }
        result.unshift(current);
    }
    return result;
};

结语

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