94. 二叉树的中序遍历

题目描述

给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。

示例 1：

  1
    \
    2
    /
  3

输入：root = [1,null,2,3]
输出：[1,3,2]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

提示：

• 树中节点数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

解题方法

方法一：dfs

/**
 * 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 inorderTraversal = function (root) {
  if (!root) {
    return [];
  }
  const res = [];
  const dfs = (node) => {
    if (node.left) {
      dfs(node.left);
    }
    res.push(node.val);
    if (node.right) {
      dfs(node.right);
    }
  };
  dfs(root);
  return res;
};

方法二：bfs

/**
 * 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 rightSideView = function (root) {
  if (!root) {
    return [];
  }
  const res = [];
  const queue = [root];
  while (queue.length) {
    let len = queue.length;
    while (len) {
      const node = queue.shift();
      if (len === 1) {
        res.push(node.val);
      }
      if (node.left) {
        queue.push(node.left);
      }
      if (node.right) {
        queue.push(node.right);
      }
      len--;
    }
  }
  return res;
};

/**
 * 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 rightSideView = function (root) {
  if (!root) {
    return [];
  }
  const res = [];
  const queue = [root];
  while (queue.length) {
    let len = queue.length;
    for (let i = 0; i < len; i += 1) {
      const node = queue.shift();
      if (i === len - 1) {
        res.push(node.val);
      }
      if (node.left) {
        queue.push(node.left);
      }
      if (node.right) {
        queue.push(node.right);
      }
    }
  }
  return res;
};