{   node(j) + 1; 　　　　　　　　if node(j) == node(k) 　　--(1)
node(i) = {    {   max(node(j), node(k)); 　　if node(j) != node(k) 　　--(2)

#include <cstdio>#include <vector>
#include <stack>#include <functional>struct tree_node_t {int left_index;int right_index;int value;tree_node_t(int li = -1, int ri = -1, int v = -1): left_index(li), right_index(ri), value(v) {}
};class machine {
public:machine() {}// *) 输入数据并构建二叉树void init_and_buildtree(int n, const std::vector<int> &datas) {arr_tree.resize(n);for ( int i = 1; i < datas.size(); i++ ) {int idx = datas[i] - 1;if ( arr_tree[idx].left_index == -1 ) {arr_tree[idx].left_index = i;} else {arr_tree[idx].right_index = i;}}}// *) 计算结果// *) 这边采用模拟堆栈的方式, 来实现递归调用, 因为节点有10000个. // *）最差情况会导致二叉树成链表, 导致递归的调用栈达到10000int calculator() {std::stack<int> frames;frames.push(0);while ( !frames.empty() ) {int idx = frames.top();const int &li = arr_tree[idx].left_index;const int &ri = arr_tree[idx].right_index;if ( li != -1 && arr_tree[li].value == -1 ) {frames.push(li);} else if ( ri != -1 && arr_tree[ri].value == -1 ) {frames.push(ri);} else {if ( li == -1 ) {arr_tree[idx].value = 1;} else if ( li != -1 && ri == -1 ) {arr_tree[idx].value = arr_tree[li].value;} else if ( li != -1 && ri != -1 ) {if ( arr_tree[li].value == arr_tree[ri].value ) {arr_tree[idx].value = arr_tree[li].value + 1;} else {arr_tree[idx].value = std::max(arr_tree[li].value, arr_tree[ri].value);} }frames.pop();}} return arr_tree[0].value;}public:std::vector<tree_node_t> arr_tree;
};int main()
{int n;while ( scanf("%d", &n) != EOF ) {std::vector<int> datas(n, 0);for ( int i = 1; i < n; i++ ) {scanf("%d", &datas[i]);}machine instance;instance.init_and_buildtree(n, datas);printf("%d\n", instance.calculator());}return 0;}

   // *)　递归函数, 划分子问题
int evaluate(int idx) {int li = arr_tree[idx].left_index;int ri = arr_tree[idx].right_index;if ( li != -1 ) {evaluate(li);} if ( ri != -1 ) {evaluate(ri);} if ( li == -1 ) {return arr_tree[idx].value = 1;} else if ( li != -1 && ri == -1 ) {return arr_tree[idx].value = arr_tree[li].value;} else {if ( arr_tree[li].value == arr_tree[ri].value ) {arr_tree[idx].value = arr_tree[li].value + 1;} else {arr_tree[idx].value = std::max(arr_tree[li].value, arr_tree[ri].value);}}return arr_tree[idx].value;}// *) 驱动函数
int calculator() {return evaluate(0);
}