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LeetCode OJ 145. Binary Tree Postorder Traversal

时间:2016-11-06 09:34:10      阅读:206      评论:0      收藏:0      [点我收藏+]

标签:span   str   tor   its   traversal   visit   amp   des   sum   

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?

 

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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) {
    int stack[1000], top = -1, i = 0;
    int *return_array = (int*)malloc(sizeof(int) * 1000);
    struct TreeNode *visit = NULL;
    
    while(-1 != 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;
}

 

LeetCode OJ 145. Binary Tree Postorder Traversal

标签:span   str   tor   its   traversal   visit   amp   des   sum   

原文地址:http://www.cnblogs.com/YuNanlong/p/6032545.html

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