二叉树的右视图

Posted 2021-06-13 HelloCoding08

给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能 看到的节点值。

例如，给定如下二叉树: root = [ 3,  5 , 1,  6,  2, 9, 8, null, null,7 , 4 ]

返回： [3， 1， 8， 4]

#include <stdio.h>#include <stdlib.h>
int get_max(int left, int right) { return left > right ? left : right;}
typedef int ElemType;typedef struct node Node;typedef struct node* Tree;
struct node { ElemType elem; Node* lchild; Node* rchild;};//创建树Tree create_tree (FILE* fp) { Tree tree; ElemType elem; fscanf(fp, "%d", &elem); if (0 == elem) { return NULL; } else { tree = (Tree)malloc(sizeof(Node)); tree->elem = elem; tree->lchild = create_tree(fp); tree->rchild = create_tree(fp); } return tree;}//计算树的高度int max_depth (Tree tree) { if (tree == NULL) { return 0; }    //左孩子的高度与右孩子的高度的最大值加上根节点 int depth = get_max(max_depth(tree->lchild), max_depth(tree->rchild)) + 1; return depth;}
void dfs (Node* curr, int* result, int* index, int level) { //递归的终止条件判断 if (curr == NULL) { return; } //level表示的是第几层，因为是先遍历右子树，所以每一层最先遍历 if (*index == level) { result[(*index)++] = curr->elem; }    //先遍历右子树，在遍历左子树 dfs(curr->rchild, result, index, level + 1); dfs(curr->lchild, result, index, level + 1);}
int* right_side_view (Tree tree, int* size) { int depth = max_depth(tree); int* result = (int*)malloc(sizeof(int) * depth); int index = 0; int level = 0; dfs(tree, result, &index, level); *size = index; return result;}
int main () { char path[] = "./demo.txt"; FILE* fp = fopen(path, "r"); if (fp == NULL) { perror("fopen error"); exit(-1); } Tree tree = create_tree(fp); fclose(fp); int size = 0; int* result = right_side_view(tree, &size);  for (int i = 0; i < size; i++) { printf("%d\t", result[i]); } printf("\n"); return 0; }