二叉树遍历全攻略：3种方式轻松掌握

二叉树遍历算法详解：前序、中序与后序遍历

二叉树的遍历是数据结构与算法中的基础内容，掌握前序、中序和后序遍历对于理解树结构至关重要。以下是三种遍历方式的实现方法与应用场景分析。

递归实现

递归是最直观的遍历方式，代码简洁但需要注意栈溢出问题。

前序遍历（根-左-右）

def preorderTraversal(root):
    res = []
    def traverse(node):
        if not node:
            return
        res.append(node.val)
        traverse(node.left)
        traverse(node.right)
    traverse(root)
    return res

中序遍历（左-根-右）

def inorderTraversal(root):
    res = []
    def traverse(node):
        if not node:
            return
        traverse(node.left)
        res.append(node.val)
        traverse(node.right)
    traverse(root)
    return res

后序遍历（左-右-根）

def postorderTraversal(root):
    res = []
    def traverse(node):
        if not node:
            return
        traverse(node.left)
        traverse(node.right)
        res.append(node.val)
    traverse(root)
    return res

迭代实现

迭代法通过显式使用栈来模拟递归过程，更适合处理深度较大的树。

前序遍历迭代版

def preorderTraversal(root):
    res = []
    stack = []
    if root:
        stack.append(root)
    while stack:
        node = stack.pop()
        res.append(node.val)
        if node.right:
            stack.append(node.right)
        if node.left:
            stack.append(node.left)
    return res

中序遍历迭代版

def inorderTraversal(root):
    res = []
    stack = []
    curr = root
    while curr or stack:
        while curr:
            stack.append(curr)
            curr = curr.left
        curr = stack.pop()
        res.append(curr.val)
        curr = curr.right
    return res

后序遍历迭代版

def postorderTraversal(root):
    res = []
    stack = []
    prev = None
    while root or stack:
        while root:
            stack.append(root)
            root = root.left
        root = stack.pop()
        if not root.right or root.right == prev:
            res.append(root.val)
            prev = root
            root = None
        else:
            stack.append(root)
            root = root.right
    return res

莫里斯遍历（Morris Traversal）

该算法能在O(1)空间复杂度下完成遍历，通过临时修改树结构实现。

中序遍历莫里斯版

def inorderTraversal(root):
    res = []
    curr = root
    while curr:
        if not curr.left:
            res.append(curr.val)
            curr = curr.right
        else:
            pre = curr.left
            while pre.right and pre.right != curr:
                pre = pre.right
            if not pre.right:
                pre.right = curr
                curr = curr.left
            else:
                pre.right = None
                res.append(curr.val)
                curr = curr.right
    return res

应用场景分析

前序遍历：用于复制树结构，序列化二叉树时常用
中序遍历：二叉搜索树会产生有序序列，用于验证BST性质
后序遍历：适用于删除树节点，计算子树属性等场景

复杂度对比

| 遍历方式 | 时间复杂度 | 空间复杂度 | |----------|------------|------------| | 递归 | O(n) | O(h) | | 迭代 | O(n) | O(h) | | 莫里斯 | O(n) | O(1) |

其中n为节点数，h为树高度。实际应用中应根据具体场景选择合适方法，递归简洁但可能栈溢出，迭代稳定但代码稍复杂，莫里斯节省空间但会修改树结构。

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