51.Lowest Common Ancestor of a Binary Tree（二叉树的最小公共祖先）

Level：

  Medium

题目描述：

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the binary tree.

思路分析：

  在一个二叉树中寻找两个节点的最小公共祖先，如果根不为空时，不断从其左右子树搜索，如果两个节点都在左子树，就在左子树递归查找，如果两个节点都在右子树，就在右子树递归查找，如果一个节点在左子树，一个节点在右字数，那么当前节点就是最小公共祖先。

代码：

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||root==p||root==q)
            return root;
        //查看左子树有没有目标节点，没有就为null
        TreeNode left=lowestCommonAncestor( root.left, p,q); 
        //查看右子树有没有目标节点，没有就为null
        TreeNode right=lowestCommonAncestor( root.right, p,q);
        //都不为空说明左右子树都有目标节点，那么当前节点就是最小祖先
        if(left!=null&&right!=null)
            return root;
        //左子树为空就去右子树找，否则去左子树找
        return left!=null?left:right;
    }
}