145. 二叉树的后序遍历

题目来源：145. 二叉树的后序遍历

/**
 * 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 postorderTraversal = function(root) {
  if (root === null)
    return [];
  let cur = root;
  let res = [];
  var addPath=(node) =>{
    let arr = [];
    while(node != null){
      arr.push(node.val);
      node = node.right;
    }
    return arr.reverse();
  }
  while (cur) {
    let node = cur.left;
    if (node !== null) {
      while (node !== null && node.right !== null && node.right !== cur) {
        node = node.right;
      }
      if (node.right === null) {
        node.right = cur;
        cur = cur.left;
        continue;
      } else {
        node.right = null;
        res.push(...addPath(cur.left));
      }
    }
    cur = cur.right;
  }
  res.push(...addPath(root));
  return res;
};

给定一个二叉树，返回它的 后序 遍历。

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 postorderTraversal(self, root: TreeNode) -> List[int]:
      def postorder(node: TreeNode):
        if not node:
          return 
        postorder(node.left)
        postorder(node.right)
        res.append(node.val)
      res = list()
      postorder(root)
      return res

      
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
      res = list()
      if not root:
        return res
      stack = list()
      node  = root
      pre = None

      while stack or node:
        while node:
          stack.append(node)
          node = node.left
        node = stack.pop()

        if not node.right or node.right is pre:
          res.append(node.val)
          pre = node
          node = None
        else:
          stack.append(node)
          node = node.right

      return res
 
 
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
      def addPath(node: TreeNode):
        tmp = list()
        while node:
          tmp.append(node.val)
          node = node.right
        tmp.reverse()
        res.extend(tmp)
        
      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:
            cur.right = None
            addPath(node.left)
        node = node.right
      addPath(root)
      return res 

 

示例:

输入: [1,null,2,3]  
   1
    \
     2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？