题目:

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.

代码:

/**
* 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 || root==p || root==q ) return root;
TreeNode* l = Solution::lowestCommonAncestor(root->left, p, q);
TreeNode* r = Solution::lowestCommonAncestor(root->right, p, q);
if ( l && r )
{
return root;
}
else
{
return l ? l : r;
}
}
};

tips:

直接学习大神的思路(http://bookshadow.com/weblog/2015/07/13/leetcode-lowest-common-ancestor-binary-tree/),确实很巧妙。

总体来说,就是分叉找;终止条件是到了叶子或者找到p和q的一个就返回了。

这里有个思维误区,为啥找到p或q的一个就返回了,如果先找到p而q在p的下面呢?那就正好了,反正这俩点如果q在p的下面,那么根据定义p就是p和q的公共祖先。

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