題目描述
給你一個二叉樹,請你返回其按 層序遍歷 得到的節點值。 (即逐層地,從左到右訪問所有節點)。
示例:
二叉樹:[3,9,20,null,null,15,7],
返回其層次遍歷結果:
[
[3],
[9,20],
[15,7]
]
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/binary-tree-level-order-traversal
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白話題目:
算法:
詳細解釋關注 B站 【C語言全代碼】學渣帶你刷Leetcode 不走丟 https://www.bilibili.com/video/BV1C7411y7gB
C語言完全代碼
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
#define MAXSIZE 1000
int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes)
{
if (!root) { /* 空樹 */
*returnSize = 0;
return;
}
struct TreeNode *que[MAXSIZE], *p; /* 定義並初始化隊列 */
int front = 0, rear = 0;
int** r = malloc(sizeof(int*) * MAXSIZE); /* 申請一個二維數組存儲層次遍歷結果 */
*returnColumnSizes = malloc(sizeof(int) * MAXSIZE);
int lev = 0, curCol; /* 層號和當前層的結點個數 */
rear = (rear+1) % MAXSIZE; /* 根結點入隊 */
que[rear] = root;
while (front != rear) {
curCol = (rear - front + MAXSIZE) % MAXSIZE; /* 統計當前層的結點個數 */
(*returnColumnSizes)[lev] = curCol;
r[lev] = malloc(sizeof(int) * curCol);
for (int i = 0; i < curCol; ++i) { /* 記錄當前層的結點 */
front = (front + 1) % MAXSIZE; /* 出隊 */
p = que[front];
r[lev][i] = p->val;
if (p->left) { /* 左子樹存在,入隊 */
rear = (rear + 1) % MAXSIZE;
que[rear] = p->left;
}
if (p->right) { /* 右子樹存在,入隊 */
rear = (rear + 1) % MAXSIZE;
que[rear] = p->right;
}
}
lev++; /* 進入下一層 */
}
*returnSize = lev;
return r;
}