107. Binary Tree Level Order Traversal II

[Tcdian]

107. Binary Tree Level Order Traversal II

给定一个二叉树，返回其节点值自底向上的层次遍历。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

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]
]

[Tcdian]

Solution

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrderBottom = function(root) {
    if (root === null) {
        return [];
    }
    const result = [];
    const queue = [[root, 0]];
    while(queue.length !== 0) {
        const [node, depth] = queue.shift();
        result[depth] = [...(result[depth] || []), node.val];
        if (node.left !== null) {
            queue.push([node.left, depth + 1]);
        }
        if (node.right !== null) {
            queue.push([node.right, depth + 1]);
        }
    }
    return result.reverse();
};

• TypeScript Solution

/**
 * 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 levelOrderBottom(root: TreeNode | null): number[][] {
    if (root === null) {
        return [];
    }
    const result: number[][] = [];
    const queue: [TreeNode, number][] = [[root, 0]];
    while(queue.length !== 0) {
        const [node, depth] = queue.shift() as [TreeNode, number];
        result[depth] = [...(result[depth] || []), node.val];
        if (node.left !== null) {
            queue.push([node.left, depth + 1]);
        }
        if (node.right !== null) {
            queue.push([node.right, depth + 1]);
        }
    }
    return result.reverse();
};