leetcode.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   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.

/**
 * 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==NULL||p==NULL||q==NULL)
            return NULL;
            
        vector<TreeNode*> path1;
        getPath(root,p,path1);
        
        vector<TreeNode*> path2;
        getPath(root,q,path2);
        
        int size1 = path1.size();
        int size2 = path2.size();
        // 求最近祖先
        TreeNode* node = NULL;
        for(int i = 0,j = 0;i <= size1 && j <= size2;++i,++j){
            if((i == size1 || j == size2) || path1[i] != path2[j]){
                node = path1[i-1];
                break;
            }//if
        }//for
        return node;
       
    }
    bool getPath(TreeNode* root,TreeNode* node,vector<TreeNode*>& path)
    {
        path.push_back(root);
        if(root == node) {
            return true;
        }//if
        bool isExits = false;
        // 左子树
        if(root->left) {
            isExits = getPath(root->left,node,path);
        }//if
        // 右子树
        if(!isExits && root->right) {
            isExits = getPath(root->right,node,path);
        }//if
        if(!isExits) {
            path.pop_back();
        }//if
        return isExits;
    }
    
};