【两次过】Lintcode 88. Lowest Common Ancestor of a Binary Tree

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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。

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

注意事项

假设给出的两个节点都在树中存在

---

解题思路：

分治法。

假设左子树找到了LCA，右子树找到了LCA，说明root是LCA

假设左子树找到了LCA，右子树没找到，说明左子树LCA就是整棵树的LCA，同理右子树。

/**
 * 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: A TreeNode in a Binary.
     * @param B: A TreeNode in a Binary.
     * @return: Return the least common ancestor(LCA) of the two nodes.
     */
     //如果找到了就返回这个LCA
     //如果只找到了A，就返回A
     //如果只找到了B，就返回B
     //如果都没有返回null
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        // write your code here
        if(root == null || root == A || root == B)
            return root;
            
        TreeNode rightNode = lowestCommonAncestor(root.right, A, B);
        TreeNode leftNode = lowestCommonAncestor(root.left, A, B);

        if(rightNode != null && leftNode != null)
            return root;
            
        if(rightNode != null)
            return rightNode;
            
        if(leftNode != null)
            return leftNode;
            
        return null;
    }
}