栈的应用之二叉树的中序遍历

中序遍历

一、实例要求

• 给定一个二叉树的根节点 root ，返回 它的 中序 遍历 ；
  

二、实例分析

• 1、首先定义了二叉树的结构 TreeNode，以及栈的结构 Stack 和相关操作；
• 2、然后实现了中序遍历函数 inorderTraversal，在遍历过程中使用栈来辅助实现；
• 3、最后释放了内存；

三、示例代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
// 定义栈的结构和相关操作
struct StackNode {
    struct TreeNode *node;
    struct StackNode *next;
};

struct Stack {
    struct StackNode *top;
};

void stackPush(struct Stack *stack, struct TreeNode *node) {
    struct StackNode *newNode = (struct StackNode *)malloc(sizeof(struct StackNode));
    newNode->node = node;
    newNode->next = stack->top;
    stack->top = newNode;
}

struct TreeNode *stackPop(struct Stack *stack) {
    if (stack->top == NULL) {
        return NULL;
    }
    struct TreeNode *node = stack->top->node;
    struct StackNode *temp = stack->top;
    stack->top = stack->top->next;
    free(temp);
    return node;
}

int isStackEmpty(struct Stack *stack) {
    return (stack->top == NULL);
}
int* inorderTraversal(struct TreeNode* root, int* returnSize) {
   // 创建一个用于存储遍历结果的数组
    int *result = (int *)malloc(100 * sizeof(int));
    *returnSize = 0;
    
    // 创建一个栈
    struct Stack *stack = (struct Stack *)malloc(sizeof(struct Stack));
    stack->top = NULL;
    
    // 遍历二叉树
    while (root != NULL || !isStackEmpty(stack)) {
        // 将当前节点及其左子节点依次入栈
        while (root != NULL) {
            stackPush(stack, root);
            root = root->left;
        }
        // 弹出栈顶节点，访问该节点的值，并将其右子节点设为当前节点
        root = stackPop(stack);
        result[(*returnSize)++] = root->val;
        root = root->right;
    }
    
    // 释放栈内存
    free(stack);
    
    return result; 
}

四、运行结果