/*
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.
思路1:
1:利用二個棧來存儲兩個節點的路徑
2:使路徑長的節點,從棧頂pop節點,直到兩個棧的長度相同
3:開始從棧頂比較,如果相等,即爲最低的父節點
思路2:
兩個節點在同一個子樹或不同子樹,在同一個子樹直接返回遇到的第一個節點即可;
在不同的子樹,當兩個節點都遇到後,下一個父節點即是所求
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root) return NULL;
stack<TreeNode*> sp,sq;
findNodePath(root,q,sq);
findNodePath(root,p,sp);
while(!sq.empty() && !sp.empty()){
if(sq.size()>sp.size())
sq.pop();
else if(sq.size()<sp.size())
sp.pop();
else
break;
}
while(!sq.empty() && !sp.empty()){
if(sq.top()==sp.top())
return sq.top();
sq.pop();
sp.pop();
}
return NULL;
}
void findNodePath(TreeNode* root,TreeNode* target,stack<TreeNode*> &st){
if(!root) return;
st.push(root);
if(st.top()==target) return;
if(root->left) findNodePath(root->left,target,st);
if(st.top()!=target && root->right) findNodePath(root->right,target,st);
if(st.top()!=target) st.pop();
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root || root==p || root==q) return root;
TreeNode* left=lowestCommonAncestor(root->left,p,q);
TreeNode* right=lowestCommonAncestor(root->right,p,q);
return !left ? right : !right ? left : root;
}
};
236. Lowest Common Ancestor of a Binary Tree
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