【7】236. Lowest Common Ancestor of a Binary Tree

 

236. Lowest Common Ancestor of a Binary Tree

• Total Accepted: 82386
• Total Submissions: 280908
• Difficulty: Medium
• Contributors: Admin

 

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.

DFS:

1. Base Case: 一般是找到符合条件的

2. recursive: 调用本身

3. Judgement: 一般是排除不符合条件的

/**
 * 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) {
        //在二叉树中来搜索p和q，然后从路径中找到最后一个相同的节点即为父节点，我们可以用递归来实现
        //base case
        if(!root || root == p || root == q) return root;
        //recursive
        TreeNode* left = lowestCommonAncestor(root -> left, p, q);
        TreeNode* right = lowestCommonAncestor(root -> right, p, q);
        //judgement
        if(left && right) return root;//如果一个在左子树 一个在右子树 return root
        return left ? left : right;//如果left不为空 return left 反之返回right
    }
};