94. 二叉树的中序遍历

题目来源：94. 二叉树的中序遍历

给定一个二叉树的根节点 root ，返回它的 中序 遍历。

 

/**
 * 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 == null){
        return [];
    }
    let cur = root;
    let res = [];
    while(cur){
        if(cur.left == null){
            res.push(cur.val);
            cur = cur.right;
        }else{
            let node = cur.left;
            while(node != null && node.right != null && node.right != cur){
                node = node.right;
            }
            if(node.right == null){
                node.right = cur;
                cur = cur.left;
            }else{
                node.right = null;
                res.push(cur.val)
                cur = cur.right
            }
        }
    }
    return res;
};

示例 1：

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]

示例 4：

输入：root = [1,2]
输出：[2,1]

示例 5：

输入：root = [1,null,2]
输出：[1,2]

 Python3

from typing import List

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
      def inorder(node: TreeNode):
        if not node:
          return
        inorder(node.left)
        res.append(node.val)
        inorder(node.right)

      res = list()
      inorder(root)

      return res

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
      res = list()
      if not root:
        return res
      stack = list()
      node = root
      while not stack or node:
        while node:
          stack.append(node)
          node = node.left          
        node = stack.pop()
        res.append(node.val)
        node = node.right
      return res
      

class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:   
      res = list()
      if not root:
        return res
      node = root
      while node:
        cur = node.left
        if cur:
          while cur.right and cur.right is not node:
            cur = cur.right
          if not cur.right:
            cur.right = node
            node = node.left
            continue
          else:
            res.append(node.val)
            cur.right = None
        else:
          res.append(node.val)
        node = node.right
      return res

 

提示：

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

 

进阶: 递归算法很简单，你可以通过迭代算法完成吗？