后序遍历二叉树,非递归实现

题目

题目来源 : LeetCode
后序遍历二叉树,非递归实现

实现

//先实现栈

int* _postorderTraversal(struct TreeNode* root, int* a){
    Stack s;
    StackInit(&s);
    int index = 0;
    struct TreeNode* right = NULL;
    struct TreeNode* cur;
    if(root){
        StackPush(&s, root);
        root = root->left;
    }
    while(StackEmpty(&s) == 1 || root){
        //访问左子树
        while(root){
            StackPush(&s, root);
            root = root->left;
        }
        //右子树为空 或者 已经遍历 则访问根节点
        cur = StackTop(&s);
        if(cur->right == NULL || right == cur->right){
            a[index++] = cur->val;
            right = cur;
            StackPop(&s);
        }
        //访问右子树
        else{
            root = cur->right;
        }
    }
    return a;
}

//获取二叉树的节点个数
int Size(struct TreeNode* root){
    if(!root){
        return 0;
    }
    return Size(root->left) + Size(root->right) + 1;
}
int* postorderTraversal(struct TreeNode* root, int* returnSize){
    *returnSize = Size(root);
    int* a = (int*)malloc(sizeof(int) * (*returnSize));
    if(!root){
        return a;
    }
    a = _postorderTraversal(root, a);
    return a;
}