题目:
Lowest Common Ancestor of a Binary Tree
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______3______
/
___5__ ___1__
/ /
6 _2 0 8
/
7 4
For example, the lowest common ancestor (LCA) of nodes 5
and 1
is 3
.
Another example is LCA of nodes 5
and 4
is 5
,
since a node can be a descendant of itself according to the LCA definition.
分析:
求二叉树节点最低公共祖先的题目,应该做到下面两点:
1、时间复杂度O(n),仅仅需遍历一遍二叉树。
2、要推断输入的两个节点是否在二叉树中。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ struct node { TreeNode* res; bool bp,bq; node(TreeNode* n,bool x,bool y):res(n),bp(x),bq(y){} }; class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if(!root || !p || !q) return nullptr; return lowestCommonAncestor_core(root,p,q).res; } node lowestCommonAncestor_core(TreeNode* root, TreeNode* p, TreeNode* q) { if(root==nullptr) return node(nullptr,false,false); if(root==p) { return hasnode(root,q)?node(root,true,true):node(root,true,false); } if(root==q) { return hasnode(root,p)?node(root,true,true):node(root,false,true); } if(root->left) { auto r=lowestCommonAncestor_core(root->left,p,q); if(r.bp && r.bq) return r; else if(!r.bp && r.bq) { if(hasnode(root->right,p)) return node(root,true,true); else return node(nullptr,false,true); } else if(r.bp && !r.bq) { if(hasnode(root->right,q)) return node(root,true,true); else return node(nullptr,true,false); } } return lowestCommonAncestor_core(root->right,p,q); } bool hasnode(TreeNode* root, TreeNode* p) { if(root==p) return true; if(root==nullptr) return false; return hasnode(root->left,p) || hasnode(root->right,p); } };