LeetCode OJ 145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [3,2,1].

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

Subscribe to see which companies asked this question

解答

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* postorderTraversal(struct TreeNode* root, int* returnSize) {
    ], top = -, i = ;
    );
    struct TreeNode *visit = NULL;

     != top||NULL != root){
        while(NULL != root){
            stack[++top] = root;
            root = root->left;
        }
        root = stack[top];
        if(NULL == root->right||visit == root->right){
            return_array[i++] = root->val;
            top--;
            visit = root;
            root = NULL;
        }
        else{
            root = root->right;
        }
    }
    *returnSize = i;
    return return_array;
}
上一篇:ORACLE 11G内存管理方式


下一篇:转载文章----十步完全理解SQL