C++ — 迭代法遍历二叉树（LeetCode.94 - Binary Tree Inorder Traversal）

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree [1,null,2,3],

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

　　我的AC代码，递归写法比较简单，所以试试迭代：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nums;
        stack<TreeNode *> stackNode;
        while(root)
        {
            stackNode.push(root);
            if (root->left != NULL)
            {
                root = root->left;
                continue;
            }
            nums.push_back(root->val);
            stackNode.pop();
            if (root->right != NULL)
            {
                root = root->right;
                continue;
            }

            TreeNode * node = root;
            if (!stackNode.empty())
            {
                root = stackNode.top();
                root->left = NULL;
                stackNode.pop();
            }

            if (node == root)
                break;
        }
        return nums;
    }
};

　　利用递归栈的思路，自己维护一个栈，时间复杂度应该是O(n)，空间复杂度O(n)。

　　看看AC后的详细细节：

　　运行时间是3ms。

　　不过我也写了一个递归版的：

class Solution {
public:
    void helper(TreeNode* root, vector<int> & nums) {
        if (root) {
            helper(root->left, nums);
            nums.push_back(root->val);
            helper(root->right, nums);
        }
    }

    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> nums;
        helper(root, nums);
        return nums;
    }
};

　　这个运行结果感觉差不多：

　　完整的迭代算法：

#include <iostream>
#include <vector>
#include <stack>

using namespace std;

class Tree {
public:
	int val;
	Tree * left;
	Tree * right;
	Tree() : val(), left(NULL), right(NULL) {}
	void CreateTree(Tree * &root);
	Tree* insert(Tree * root, int x);
	void buildTree(Tree* root, vector<int> nums);
	void inorderTraversal(Tree * root);
	void iterative_inorderTraversal(Tree * root);
};

void Tree::CreateTree(Tree * &root)
{
    int val;
    cin >> val;
    if (val == -1)
        root = NULL;
    else
    {
        root = new Tree;
        root->val = val;
        CreateTree(root->left);
        CreateTree(root->right);
    }
}

Tree* Tree::insert(Tree * root, int x)
{
	Tree * node = new Tree;
	node->val = x;

	if (root == NULL)
    {
        root = node;
        return root;
    }
	else if (root->left == NULL)
    {
        root->left = node;
        return root;
    }
	else if (root->right == NULL)
    {
        root->right = node;
        return root;
    }
    else
        node->left = root;

	return node;
}

void Tree::buildTree(Tree* root, vector<int> nums)
{
	for (auto n : nums)
		root = insert(root, n);
}

void Tree::inorderTraversal(Tree * root)
{
	if (root != NULL)
	{
		inorderTraversal(root->left);
		cout << root->val << ‘ ‘;
		inorderTraversal(root->right);
	}
}

void Tree::iterative_inorderTraversal(Tree * root)
{
    stack<Tree *> stackNode;
    while(root)
    {
        stackNode.push(root);
        if (root->left != NULL)
        {
            root = root->left;
            continue;
        }
        cout << root->val << ‘ ‘;
        stackNode.pop();
        if (root->right != NULL)
        {
            root = root->right;
            continue;
        }

        Tree * node = root;
        if (!stackNode.empty())
        {
            root = stackNode.top();
            root->left = NULL;
            stackNode.pop();
        }

        if (node == root)
            break;
    }
}

int main()
{
	//vector<int> nums{ 1,2,3 };
	Tree * root = NULL;
	root->CreateTree(root);

	//root->buildTree(root, nums);
	root->inorderTraversal(root);
	cout << endl;
    root->iterative_inorderTraversal(root);

    return 0;
}

原文地址：https://www.cnblogs.com/darkchii/p/8576029.html