1123. 最深叶节点的最近公共祖先
就找出最后一层的所有节点,依次求lca就行
lca暴力就行
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: int height; int idx[1010]; int pre[1010]; vector<TreeNode*> v; void dfs1(TreeNode* root, int h, int f) { if(root == nullptr) return; height = max(height, h); idx[root->val] = h; pre[root->val] = f; dfs1(root->left, h + 1, root->val); dfs1(root->right, h + 1, root->val); } void dfs2(TreeNode* root, int h) { if(root == nullptr) return; if(h == height) { v.push_back(root); } dfs2(root->left, h + 1); dfs2(root->right, h + 1); } int query(int x, int y) { int f1 = idx[x], f2 = idx[y]; while(x != y) { while(x != y && f1 >= f2) { x = pre[x]; f1 = idx[x]; } while(x != y && f1 <= f2) { y = pre[y]; f2 = idx[y]; } } return x; } TreeNode* dfs3(TreeNode* root, int x) { if(root == nullptr) return nullptr; if(root->val == x) return root; TreeNode* r1 = dfs3(root->left, x); if(r1) return r1; TreeNode* r2 = dfs3(root->right, x); if(r2) return r2; return nullptr; } TreeNode* lcaDeepestLeaves(TreeNode* root) { height = 0; dfs1(root, 0, -1); dfs2(root, 0); int len = v.size(); int x = v[0]->val, y; for(int i = 1; i < len; i++) { y = v[i]->val; x = query(x, y); } return dfs3(root, x); } };
自己选择的路,跪着也要走完。朋友们,虽然这个世界日益浮躁起来,只要能够为了当时纯粹的梦想和感动坚持努力下去,不管其它人怎么样,我们也能够保持自己的本色走下去。