算法day 14 第六章二叉树

今日内容：

• 理论基础

• 递归遍历

• 迭代遍历

• 统一迭代

递归遍历 （必须掌握）

三个要点：

• 参数以及返回值
• 终止条件
• 单层循环逻辑

前序遍历：中左右

中序遍历：左中右

后序遍历：左右中

• 144.二叉树的前序遍历(opens new window)

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result =[]
        def Traversal(root):
            if root == None:
                return
            result.append(root.val)
            Traversal(root.left)
            Traversal(root.right)
            
        Traversal(root)
        return result

• 145.二叉树的后序遍历(opens new window)

  class Solution(object):
      def postorderTraversal(self, root):
          """
          :type root: TreeNode
          :rtype: List[int]
          """
          result = []
   
          def traversal(root):
              if root == None:
                  return
              traversal(root.left)    # 左
              traversal(root.right)   # 右
              result.append(root.val) # 后序
   
          traversal(root)
          return result
  

中右左-反转变成左右中

• var postorderTraversal = function(root) {
      // 左右中
      let result = []
      let stack = []
      stack.push(root)
      while(stack.length){
          let node = stack.pop()
          if(node){
              result.push(node.val)
              if(node.left){
              stack.push(node.left)
              }
              if(node.right){
                  stack.push(node.right)
              }
          }   
      }
      return result.reverse()
  };
  

• 94.二叉树的中序遍历

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        #中序遍历:左中右
        result = []
        def traversal(root):
            if root is None:
                return
            traversal(root.left)
            result.append(root.val)
            traversal(root.right)
        
        traversal(root)
        return result

递归法

var inorderTraversal = function(root) {
    let result =[]
    const traversal = function(node){
        // 左中右
        if(node === null){
            return 
        }
        if(node.left){
            traversal(node.left)
        }
        result.push(node.val)
        if(node.right){
            traversal(node.right)
        }
        return result
    }
    if(root === null){
        return []
    }
    return traversal(root)
    
};

迭代法

var inorderTraversal = function(root) {
    let result = []
    let stack = []
    let cur = root
    while(stack.length||cur){
        if(cur){
            stack.push(cur)
            cur = cur.left
        }else{
            cur = stack.pop()
            result.push(cur.val)
            cur = cur.right
        }
    }
    return result
};

题目链接/文章讲解/视频讲解：代码随想录

迭代遍历/统一迭代遍历 （基础不好的录友，迭代法可以放过）

前序改为后序只需要三行代码：

中左右->中右左->左右中

访问节点/处理节点

#前序
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        stack = []
        result = []
        if root is not None:
            stack.append(root)
        while stack:
            node = stack.pop()
            if node is not None:
                if node.right is not None:
                    stack.append(node.right)
                if node.left is not None:
                    stack.append(node.left)
                stack.append(node)
                stack.append(None)
            else:
                node = stack.pop()
                result.append(node.val)
        return result

#后序
class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result = []
        stack = []
        if root:
            stack.append(root)
        while stack:
            node = stack.pop()
            if node != None:
               
                if node.left:
                    stack.append(node.left)
                if node.right:
                    stack.append(node.right)
                stack.append(node)
                stack.append(None)
            else:
                node = stack.pop()
                result.append(node.val)
        return reversed(result)

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
#中序遍历
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        #中序遍历:左中右
        result = []
        stack = []
        if root:
            stack.append(root)
        while stack:
            node = stack.pop()
            if node is not None:
                if node.right is not None:
                    stack.append(node.right)
                    
                stack.append(node)
                stack.append(None)
                
                if node.left is not None:
                    stack.append(node.left)
            else:
                node = stack.pop()
                result.append(node.val)
        return result

如何标记呢，就是要处理的节点放入栈之后，紧接着放入一个空指针作为标记。 这种方法也可以叫做标记法。

题目链接/文章讲解/视频讲解：代码随想录，代码随想录