PAT 甲级 1151 LCA in a Binary Tree (30 分) PAT 甲级 1151 LCA in a Binary Tree 思路: 代码: 太惭愧了,这题暴力解的,玩不来各个大佬的lca算法。 思路: 1.将中根先根序列存储,然后用递归分治分别求出每个key的父结点,同时记录结点层次; 2.用全局变量存储中根先根序列,然后以下标范围来限定子树范围...
For each given pair of U and V, print in a lineLCA of U and V is A.if the LCA is found andAis the key. But ifAis one of U and V, printX is an ancestor of Y.whereXisAandYis the other node. If U or V is not found in the binary tree, print in a lineERROR: U is no...
//lca 递归之美 parms:中序左边界 中序右边界 先序遍历到的根的下标 ab值 void lca(intinl,intinr,intpreRoot,inta,intb) {if(inl > inr)return;//出错intinRoot =pos[pre[preRoot]], aIn =pos[a], bIn =pos[b];//取当前根、左边、右边、位置if(aIn < inRoot && bIn < inRoot) lca(inl...
printf("LCA of %d and %d is %d.\n",a,b,pre[preRoot]); } elseif(inRoot > aIn&&inRoot > bIn) {//a,b都在跟的左边,则继续遍历左子树 lca(inl,inRoot-1,preRoot+1,a,b); } elseif(inRoot < aIn&&inRoot < bIn) { lca(inRoot+1,inr,preRoot+1+(inRoot-inl),a,b); } elsei...
1151 LCA in a Binary Tree C++版 1.题意 给出中序和先序遍历序列,让你找出给定的查询记录的最小父节点,如果给定的值不存在,则按照相应的格式输出。指定的格式如下: For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if...
1151 LCA in a Binary Tree C++版 1.题意 给出中序和先序遍历序列,让你找出给定的查询记录的最小父节点,如果给定的值不存在,则按照相应的格式输出。指定的格式如下: For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the ke... ...
[i]为当前结点i的父节点int n,m,in[MAX];//确定每个结点的父节点和深度void createTree(int root,int left,int right,int father,int level){if(left>right)return;int i=left;while(i<=right && in[i]!=pre[root].data)++i;//找到根结点在中序遍历序列中的位置ipre[root]=Node(pre[root].data...
voidcreateTree(introot,intleft,intright,intfather,intlevel){ if(left>right)return; pre[root]={pre[root].key,father,level}; inti=left; while(i<=right&&pre[root].key!=in[i]) i++; createTree(root+1,left,i-1,root,level+1); ...
Given any two nodes in a binary tree, you are supposed to find their LCA. Input Specification: Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the nu...
=nil{returnroot}// Otherwise, LCA is either in left subtree or right subtreeifleftLCA!=nil{returnleftLCA}returnrightLCA}funcmain(){// Example usage:// Construct a binary tree// 3// / \// 5 1// / \ / \// 6 2 0 8// / \// 7 4root:=&TreeNode{Val:3}root.Left=&TreeNode...