LeetCode OJ：Lowest Common Ancestor of a Binary Tree（最近公共祖先）

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.

找最近的公共祖先，仔细想一想，其实l与r的最小的公共祖先c满足一定条件，那就是l以及r一定在c的左右分支上，不可能都是左或右分支的。否则一定就不是最近的公共祖先，

代码如下，用递归写出来还是比较简单易懂的。

 /**
  * 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) {
         if (root == NULL) return  NULL; //无其他节点，直接返回
         if (root == p || root == q) return root;
         TreeNode * leftNode = lowestCommonAncestor(root->left, p, q);
         TreeNode * rightNode = lowestCommonAncestor(root->right, p, q);
         if (leftNode && rightNode) return root;　　//找到LCA，返回LCA
         return leftNode ? leftNode : rightNode;　　
     }
 };

还有一种方法是遍历tree，然后找出到达p以及q分别的路径，找到路径之后，遍历两条路径，出现分叉的第一个点就是p与q的LCA。具体代码先不贴了  比较麻烦。

java版本的如下所示，方法相同：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null)
            return null;
        if(root == p || root == q)//切断路径，不再向下执行了
            return root;
        TreeNode leftNode = lowestCommonAncestor(root.left, p, q);
        TreeNode rightNode = lowestCommonAncestor(root.right, p, q);
        if(leftNode != null && rightNode != null)
            return root;
        if(leftNode!=null) return leftNode;
        if(rightNode != null) return rightNode;
        return null;
    }
}