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...
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, inRoot -1, preRoot +1, a, b);//如果a和b都在左子树,递归查找左子树。
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 not found.orERROR: V is not found.orERROR: U and V are not found..
if(inl > inr)return; intinRoot = pos[pre[preRoot]],aIn = pos[a],bIn = pos[b];//当前根节点在中序中的位置,a在中序中的位置,b在中序中的位置 if((inRoot > aIn&&inRoot < bIn)||(inRoot < aIn&&inRoot > bIn)) { printf("LCA of %d and %d is %d.\n",a,b,pre[preRoot]);...
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... ...
1151 LCA in a Binary Tree (30 point(s)) The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. Given any two nodes in a binary tree, you are supposed to find their LCA. ...
//确定当前子树根结点的父结点和深度createTree(root+1, left,i-1, root, level+1);//递归处理左子树createTree(root+1+i-left, i+1,right, root, level+1);}int main(){int m,n,a,b;scanf("%d %d",&m,&n);//测试m对结点,树共n个结点for(int i=0;i<n;i++){scanf("%d",&in[i]...
The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants. 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 fir...
(self, root, p, q): """ :type root: TreeNode :type p: TreeNode :type q: TreeNode :rtype: TreeNode """ pathP, pathQ = self.findpath(root, p), self.findpath(root, q) if pathP and pathQ: length = min(len(pathQ), len(pathP)) ans = None for i in range(length): ...