236. 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   4For 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:利用二个栈来存储两个节点的路径2：使路径长的节点，从栈顶pop节点，直到两个栈的长度相同3：开始从栈顶比较，如果相等，即为最低的父节点思路2：    两个节点在同一个子树或不同子树，在同一个子树直接返回遇到的第一个节点即可；    在不同的子树，当两个节点都遇到后，下一个父节点即是所求*//** * 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;    }};