72 lines
1.6 KiB
C++
72 lines
1.6 KiB
C++
class Solution {
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public:
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/**
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* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
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*
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* 求二叉树的右视图
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* @param preOrder int整型vector 先序遍历
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* @param inOrder int整型vector 中序遍历
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* @return int整型vector
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*/
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TreeNode* build(
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const vector<int>& pre, int preL, int preR,
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const vector<int>& in, int inL, int inR)
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{
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if (preL > preR || inL > inR) return nullptr;
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TreeNode* root = new TreeNode(pre[preL]);
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int Search = find(in.begin() + inL, in.begin() + inR + 1, pre[preL]) - in.begin();
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int k = Search - inL; // 左子树节点数
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root->left = build(
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pre,
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preL + 1,
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preL + k,
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in,
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inL,
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Search - 1
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);
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root->right = build(
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pre,
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preL + k + 1,
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preR,
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in,
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Search + 1,
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inR
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);
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return root;
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}
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vector<int> solve(vector<int>& preOrder, vector<int>& inOrder) {
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TreeNode* root = build(
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preOrder, 0, preOrder.size() - 1,
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inOrder, 0, inOrder.size() - 1
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);
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vector<int> result;
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if (!root) return result;
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queue<TreeNode*> q;
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q.push(root);
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while (!q.empty()) {
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int sz = q.size(); // 当前层节点数
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for (int i = 0; i < sz; i++) {
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TreeNode* t = q.front(); q.pop();
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// 当前层最后一个节点
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if (i == sz - 1) {
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result.push_back(t->val);
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}
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if (t->left) q.push(t->left);
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if (t->right) q.push(t->right);
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}
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}
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return result;
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}
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}; |