【数据结构与算法】二叉树中从每个叶子结点到根结点的路径

题目

Qestion: 输出二叉树中从每个叶子结点到根结点的路径

数据结构与定义

#include <stdio.h>
#include <stdlib.h>typedef struct TreeNode
{int val;struct TreeNode *left;struct TreeNode *right;
} TreeNode;

二叉树形状

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核心代码

void LeafToRoot(TreeNode *node, int length, int *Path)
{// 结点不存在if (node == NULL)return;// 结点存在else{Path[length] = node->val;length = length + 1;// 该结点为叶子结点if (node->left == NULL && node->right == NULL){// 输出路径上每个结点的值for (int i = 0; i < length; i++){printf("%d ", Path[i]);}printf("\n");}else{LeafToRoot(node->left, length, Path);LeafToRoot(node->right, length, Path);}}
}

核心代码快照

在这里插入图片描述

全部代码

#include <stdio.h>
#include <stdlib.h>typedef struct TreeNode
{int val;struct TreeNode *left;struct TreeNode *right;
} TreeNode;void inputTree(TreeNode *root)
{root->val = 1;root->left = (TreeNode *)malloc(sizeof(TreeNode));root->left->val = 2;root->left->left = (TreeNode *)malloc(sizeof(TreeNode));root->left->right = (TreeNode *)malloc(sizeof(TreeNode));root->left->left->val = 4;root->left->left->left = NULL;root->left->left->right = NULL;root->left->right->val = 5;root->left->right->left = NULL;root->left->right->right = NULL;root->right = (TreeNode *)malloc(sizeof(TreeNode));root->right->val = 3;root->right->left = (TreeNode *)malloc(sizeof(TreeNode));root->right->left->val = 6;root->right->left->left = (TreeNode *)malloc(sizeof(TreeNode));root->right->left->left->val = 8;root->right->left->left->left = NULL;root->right->left->left->right = NULL;root->right->left->right = NULL;root->right->right = (TreeNode *)malloc(sizeof(TreeNode));root->right->right->val = 7;root->right->right->left = NULL;root->right->right->right = NULL;
}void LeafToRoot(TreeNode *node, int length, int *Path)
{// 结点不存在if (node == NULL)return;// 结点存在else{Path[length] = node->val;length = length + 1;// 该结点为叶子结点if (node->left == NULL && node->right == NULL){// 输出路径上每个结点的值for (int i = 0; i < length; i++){printf("%d ", Path[i]);}printf("\n");}else{LeafToRoot(node->left, length, Path);LeafToRoot(node->right, length, Path);}}
}
int main()
{int Path[20];int length = 0;TreeNode *root = new TreeNode;inputTree(root);LeafToRoot(root, length, Path);return 0;
}