最近公共祖先 · Lowest Common Ancestor

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[抄题]:

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

 “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.

[思维问题]:

 不知道子节点怎么用dc。直接对给出的p,q节点进行操作即可。

[一句话思路]:

左右分开 谁不空返回谁

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

(left != null && right != null) 时,返回的是root节点的结果,不需要再做递归运算了。是一个“合”的过程

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[总结]:

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[英文数据结构,为什么不用别的数据结构]:

只有dc算法,没有数据结构

[其他解法]:

自己写traverse函数:不好,会形成全局变量

[Follow Up]:

有parent指针的:用对齐的方法做

[LC给出的题目变变变]:

Lowest Common Ancestor of a Binary Search Tree 一模一样的,约束条件没用,直接套。

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.
     */
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
        if (root == null || A == root || B == root) {//
            return root;
        }
        //divide
        TreeNode left = lowestCommonAncestor(root.left, A, B);
        TreeNode right = lowestCommonAncestor(root.right, A, B);
        
        //conquer
        if (left != null && right != null) {
            return root;//
        }
        else if (left != null) {
            return left;
        }
        else if (right != null) {
            return right;
        }
        else {
            return null;
        }
    }
}
View Code

 

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