描述
ENG
给定一棵二叉树,找到两个节点的最近公共父节点(LCA)。
最近公共祖先是两个节点的公共的祖先节点且具有最大深度。
假设给出的两个节点都在树中存在
样例
样例 1:
输入:{1},1,1
输出:1
解释:
二叉树如下(只有一个节点):
1
LCA(1,1) = 1
样例 2:
输入:{4,3,7,#,#,5,6},3,5
输出:4
解释:
二叉树如下:
4
/ \
3 7
/ \
5 6
LCA(3, 5) = 4
思路:从根节点开始遍历,如果p和q中的任一个和root匹配,那么root就是最低公共祖先。 如果都不匹配,则分别递归左、右子树,如果有一个 节点出现在左子树,并且另一个节点出现在右子树,则root就是最低公共祖先. 如果两个节点都出现在左子树,则说明最低公共祖先在左子树中,否则在右子树。
#include <stdio.h> #include <stdlib.h>
/** * * Definition for a binary tree node. */ struct btree { int val; struct btree *left; struct btree *right; };
struct btree *lowest_common_ancestor(struct btree *root, struct btree *p, struct btree *q) { struct btree *left = NULL; struct btree *right = NULL;
if (root == NULL) { return NULL; }
if (root->val == p->val || root->val == q->val) { return root; }
left = lowest_common_ancestor(root->left, p, q); right = lowest_common_ancestor(root->right, p, q); if (left != NULL && right != NULL) { return root; }
if (left == NULL) { return right; }
return left; }
//preorder create binary tree struct btree *create_btree(int **data) { struct btree *root = NULL; int *p = *data; int i;
if(*p == 0xFF) { return NULL; } else { root = (struct btree *)malloc(sizeof(struct btree));
root->val = *p; p = ++(*data); root->left = create_btree(data); p = ++(*data); root->right = create_btree(data);
return root; } }
//preorder parse binary tree void parse_btree(struct btree *tree) { if (tree == NULL) { printf("0xFF "); return; } else { printf("%d ", tree->val); parse_btree(tree->left); parse_btree(tree->right); } }
void find_node(struct btree *tree, int val, struct btree **result) { if (tree == NULL) { return; } else { if (val == tree->val) { *result = tree; return; } find_node(tree->left, val, result); find_node(tree->right, val, result); } }
int main() { struct btree *tree = NULL; int data[13] = {35, 13, 0, 0xFF, 0xFF, -15, 0xFF, 19, 0xFF, 0xFF, -7, 0xFF, 0xFF}; int *p = data; int i; struct btree *t1 = NULL, *t2 = NULL, *result = NULL;
for (i=0; i<13; i++) { if (data[i] == 0xFF) printf("0x%X ", data[i]); else printf("%d ", data[i]); }
printf("\n"); tree = create_btree(&p); printf("\n"); parse_btree(tree); printf("\n"); find_node(tree, -15, &t1); if (t1) printf("%d, t1:%p\n", t1->val, t1); find_node(tree, 19, &t2); if (t2) printf("%d, t2:%p\n", t2->val, t2);
result = lowest_common_ancestor(tree, t1, t2); if (result) printf("%d, result:%p\n", result->val, result); }