标签:res script highlight front eve 队列 asc 维数 层次遍历
题目描述: 给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)
例如:
给定二叉树 [3,9,20,null,null,15,7],返回其自底向上的层次遍历为: [ [15,7], [9,20], [3] ]
//JS var levelOrderBottom = function(root) { let res = []; let dfs = (node, depth) => { if(!node) return; res[depth] = res[depth] || []; res[depth].push(node.val); dfs(node.left, depth + 1); dfs(node.right, depth + 1); } dfs(root, 0); return res.reverse(); };
//C语言 //单队列 //先计算二叉树的高度,得到返回二维数组长度,再正常遍历二叉树,将每层的结点数组从后往前放到结果数组中 #define MAXSIZE 2000 int maxDepth(struct TreeNode* root){ if(root == NULL) return 0; int lDepth = maxDepth(root -> left), rDepth = maxDepth(root -> right); return 1 + (lDepth > rDepth ? lDepth : rDepth); } int** levelOrderBottom(struct TreeNode* root, int* returnSize, int** returnColumnSizes){ *returnSize = maxDepth(root); if(*returnSize == 0) return NULL; int **res = (int **)malloc(sizeof(int *) * (*returnSize)); *returnColumnSizes = (int **)malloc(sizeof(int *) * (*returnSize)); struct TreeNode *queue[MAXSIZE], *p; int rear = 0, front = 0, idx = *returnSize; queue[rear++] = root; while(front != rear){ int len = rear - front; res[--idx] = (int *)malloc(sizeof(int) * len); (*returnColumnSizes)[idx] = len; for(int i = 0; i < len; i++){ p = queue[front++]; res[idx][i] = p -> val; if(p -> left) queue[rear++] = p -> left; if(p -> right) queue[rear++] = p -> right; } } return res; } //JS var levelOrderBottom = function(root) { if(!root) return []; let queueArr = [], result = [], node = null; queueArr.push(root); result.push([root.val]); while(queueArr.length != 0) { let len = queueArr.length, tmpArr = []; for(let i = 0; i < len; i++){ node = queueArr.shift(); if(node.left) { tmpArr.push(node.left.val); queueArr.push(node.left); } if(node.right) { tmpArr.push(node.right.val); queueArr.push(node.right); } } if(tmpArr.length != 0) result.push(tmpArr); } return result.reverse(); };
标签:res script highlight front eve 队列 asc 维数 层次遍历
原文地址:https://www.cnblogs.com/JesseyWang/p/13111760.html