leetcode 226. Invert Binary Tree | DFS 递归转迭代

Description

Invert a binary tree.

4
/ \
2 7
/ \ / \
1 3 6 9
to
4
/ \
7 2
/ \ / \
9 6 3 1
Trivia:
This problem was inspired by this original tweet by Max Howell:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

My close solution

class Solution {public:    TreeNode *invertTree(TreeNode *root) {        if(root==NULL) return NULL;        root->left = invertTree(root->right);        root->right = invertTree(root->left);        return root;    }};

上述代码基本思路正确, 但是写法上犯了swap错误写法的低级错误.

Discuss

• C++ 递归

TreeNode* invertTree(TreeNode* root) {    if (root) {        invertTree(root->left);        invertTree(root->right);        std::swap(root->left, root->right);    }    return root;}

• Java 递归

public TreeNode invertTree(TreeNode root) {    if (root == null) {        return null;    }    TreeNode right = invertTree(root.right);    TreeNode left = invertTree(root.left);    root.left = right;    root.right = left;    return root;}

上述两种写法基本思路是一致的, 尤其是java版本, 跟我的solution只差一个显示的swap正确写法.

• C++ stack dfs

TreeNode* invertTree(TreeNode* root) {    std::stack<TreeNode*> stk;    stk.push(root);    while (!stk.empty()) {        TreeNode* p = stk.top();        stk.pop();        if (p) {            stk.push(p->left);            stk.push(p->right);            std::swap(p->left, p->right);        }    }    return root;}

递归改迭代

二叉树先序遍历递归算法伪码

void PreorderRecursive(Bitree root){  if (root) {    /// 先visit    visit(root);    // 再压入子树    PreorderRecursive(root->lchild);     PreorderRecursive(root->rchild);   }}

二叉树先序遍历非递归伪码

void PreorderNonRecursive(Bitree root){  stack stk;  stk.push(root);  while(!stk.empty()){    p = stk.top();    /// 先对p进行处理!!!!!    visit(p);    stk.pop();    /// 再压入子树    /// 注意这里先压入的right, 这样子才是和上面等价的    if(p.rchild) stk.push(stk.rchild);    if(p.lchild) stk.push(stk.lchild);  }}

自己修改递归为stack形式, 与参考略有不同

/// 修改递归为stack的迭代形式class Solution {public:    TreeNode *invertTree(TreeNode *root) {        if(!root) return root;        stack<TreeNode *> s;        TreeNode *cur;        s.push(root);        while (!s.empty()) {            cur = s.top();            s.pop();            /// 1. 注意这里先处理cur, 即visit(cur)的所有应该做的处理            TreeNode *left = cur->left;            TreeNode *right = cur->right;            cur->left = right;            cur->right = left;            /// 2. cur遍历过了, 压入两个子树            if (cur->right) s.push(cur->right);            if (cur->left) s.push(cur->left);        }        return root;    }};

Reference

-leetcode 226

-递归算法转换为非递归算法的技巧