Tcdian / keep

今天不想做,所以才去做。
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103. Binary Tree Zigzag Level Order Traversal #267

Open Tcdian opened 4 years ago

Tcdian commented 4 years ago

103. Binary Tree Zigzag Level Order Traversal

给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。

Example

给定二叉树 [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]
Tcdian commented 4 years ago

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 zigzagLevelOrder = function(root) {
    if (root === null) {
        return [];
    }
    const result = [];
    const queue = [[root, 0]];
    while(queue.length !== 0) {
        const [node, level] = queue.shift();
        result[level] = result[level] || [];
        if (level % 2 === 0) {
            result[level].push(node.val);
        } else {
            result[level].unshift(node.val);
        }
        if (node.left !== null) {
            queue.push([node.left, level + 1]);
        }
        if (node.right !== null) {
            queue.push([node.right, level + 1]);
        }
    }
    return result;
};
/**
 * 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 zigzagLevelOrder(root: TreeNode | null): number[][] {
    if (root === null) {
        return [];
    }
    const result: number[][] = [];
    const queue: [TreeNode, number][] = [[root, 0]];
    while(queue.length !== 0) {
        const [node, level] = queue.shift() as [TreeNode, number];
        result[level] = result[level] || [];
        if (level % 2 === 0) {
            result[level].push(node.val);
        } else {
            result[level].unshift(node.val);
        }
        if (node.left !== null) {
            queue.push([node.left, level + 1]);
        }
        if (node.right !== null) {
            queue.push([node.right, level + 1]);
        }
    }
    return result;
};