求二叉树中两节点的最小公共父节点

发表于 | 分类于 | | 阅读次数:

165646729

思路：

采用递归，深度优先遍历，找到一个节点时，返回，逐层记录遍历方向，另一个节点
同，这样深度优先遍历后，可以找到这两个节点由根节点访问的路径，然后沿着这个
路径找到分叉的地方就行了。

代码：

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#include <stdlib.h>  
#include <stdio.h>  

typedef struct TNode  
{  
    int data;  
    int pathA;  
    int pathB;  
    TNode* lchild;  
    TNode* rchild;  
}TNode,*BTree;  

//采用递归进行深度优先遍历  
//计算出路径  
int traverse(BTree t,int x,int flag)  
{  
    if (t == NULL) return 0;  
    if (t->data == x) return 1;  
    if (traverse(t->lchild,x,flag)==1)  
    {  
        if (flag == 1) t->pathA = 1;  
        else t->pathB = 1;  
        return 1;  
    }  
    if (traverse(t->rchild,x,flag)==1)  
    {  
        if (flag == 1) t->pathA = 2;  
        else t->pathB = 2;  
        return 1;  
    }     
}  

int main()  
{  
    BTree tree = (BTree)malloc(sizeof(TNode));  
    tree->data = 1;  
    tree->pathA = 0;  
    tree->pathB = 0;  
    tree->lchild = (TNode*)malloc(sizeof(TNode));  
    tree->lchild->data = 2;  
    tree->lchild->pathA = 0;  
    tree->lchild->pathB = 0;  
    tree->rchild = (TNode*)malloc(sizeof(TNode));  
    tree->rchild->data = 3;  
    tree->rchild->pathA = 0;  
    tree->rchild->pathB = 0;  
    tree->rchild->lchild = NULL;  
    tree->rchild->rchild = NULL;  
    tree->lchild->lchild = (TNode*)malloc(sizeof(TNode));  
    tree->lchild->lchild->data = 4;  
    tree->lchild->lchild->pathA = 0;  
    tree->lchild->lchild->pathB = 0;  
    tree->lchild->lchild->lchild = NULL;  
    tree->lchild->lchild->rchild = NULL;  
    tree->lchild->rchild = (TNode*)malloc(sizeof(TNode));  
    tree->lchild->rchild->data = 5;  
    tree->lchild->rchild->pathA = 0;  
    tree->lchild->rchild->pathB = 0;  
    tree->lchild->rchild->lchild = NULL;  
    tree->lchild->rchild->rchild = NULL;  

    int a,b;  
    printf("please input a,b:");  
    scanf("%d %d",&a,&b);     
    if (traverse(tree,a,1)!=1) printf("no a");  
    if (traverse(tree,b,2)!=1) printf("no b");  

    //从根节点开始向下遍历，找到路径分叉点  
    TNode* node = tree;  
    while(node&&(node->pathA==node->pathB))  
    {  
        if (node->pathA == 1) node = node->lchild;  
        else node = node->rchild;  
    }     
    printf("the common father node is:%d",node->data);  
    return 0;  
}

示例：

please input a,b:4 3
the common father node is:1