LintCode：M-最近祖先

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LinCode链接

给定一棵二叉树，找到两个节点的最近公共父节点(LCA)。

最近公共祖先是两个节点的公共的祖先节点且具有最大深度。

注意事项

假设给出的两个节点都在树中存在

您在真实的面试中是否遇到过这个题？

Yes

样例

对于下面这棵二叉树

  4 / \3   7   / \  5   6

LCA(3, 5) = 4

LCA(5, 6) = 7

LCA(6, 7) = 7

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of the binary search tree.     * @param A and B: two nodes in a Binary.     * @return: Return the least common ancestor(LCA) of the two nodes.     */    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {        // write your code here        return findNode(root, A, B);    }        TreeNode findNode(TreeNode root, TreeNode A, TreeNode B){        if(root == null) return null;                //找到了目标node，返回目标        if(root.val == A.val) return root;        if(root.val == B.val) return root;                TreeNode left = findNode(root.left, A, B);        TreeNode right = findNode(root.right, A, B);                if(left!=null && right!=null)//左右两边都找到了目标，则该节点是最近LCA            return root;                return left!=null?left:right;    }}

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