Lowest Common Ancestor****

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.

 

Analyse:

Trial 1: For the current node, if p and q are at different subtree, then return current node. If p and q are both at left subtree, then recursively judge the left subtree. If p and q are both at right subtree, then recursively judge the right subtree. 

To judge the rough position of p and q, we need a function which can determine whether the current root reaches to a target node.

For example, if we want to find the LCA of node 7 and node 8. We first find that node 7 and node 5 are connected, and node 8 and node 1 are connected. That is to say, node 7 and node 8 are at the left subtree and right subtree of the root respectively. Then we return node 3.

Runtime: TIME LIMIT EXCEEDED...

/**
 * 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;
        if(reach(root->left, p) && reach(root->left, q)) 
            return lowestCommonAncestor(root->left, p, q);
        if(reach(root->right, p) && reach(root->right, q)) 
            return lowestCommonAncestor(root->right, p, q);
        return root;
    }
    bool reach(TreeNode* root, TreeNode* target){//find whether root reaches the target node
        if(!root) return false;
        if(root == target) return true;
        return reach(root->left, target) || reach(root->right, target);
    }
};

 

Trial 2: If the current node equals to one of the node, then we only need to find the location of another node. See from Solution

Runtime: 24ms.

/**
 * 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;
        if(root == p || root == q) return root;
        TreeNode* leftSub = lowestCommonAncestor(root->left, p, q);
        TreeNode* rightSub = lowestCommonAncestor(root->right, p, q);
        
        if(leftSub && rightSub) return root;
        return leftSub ? leftSub : rightSub;
    }
};