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).”
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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.
思路：如果left为空，则结果是right。若left为非空，而right为空，则结果在left。最后若两者皆不为空，则必定是p和q分别在两个子树里，不然不可能都返回了非NULL，此时直接返回root。
1 class Solution {
2 public:
3 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
4 if (!root || root == p || root == q) return root;
5 TreeNode *left = lowestCommonAncestor(root->left, p, q);
6 if (left && left != p && left != q) return left;//answer in left
7 TreeNode *right = lowestCommonAncestor(root->right, p, q);
8 if (!left) return right;//left is null, so answer depends on right
9 else if (!right) return left;//right is null, so answer in left
10 else return root;//both are not null, so root is the answer
11 }
12 };