二叉树的最近公共祖先--递归解法

 

来源: https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/solution/er-cha-shu-de-zui-jin-gong-gong-zu-xian-by-leetc-2/

信息分类： IT/算法

目标用户： 程序员/java

相关知识点：二叉树、递归

 

算法关键源码解析：

 

 

 

 

如下图，是构造的一颗二叉树，现在需要找出节点N7,节点N8的最近公共祖先。

 

                  3
     5             1
6        2      0      8
      7     4

 

以上算法中采用的是递归算法，递归算法写起来代码简洁，但没那么直观，理解起来比较费劲，下面用对具体的代码块进行了标号，

如下嵌套表展现代码的执行过程，有助有帮助大家理解递归的执行过程。

 

 

 

 

 

1

N3

2

T

1

N5

2

F

1

N6

2

F

1	NULL
2	-
3	-
4	-
5	-

 

3

F

1	NULL
2	-
3	-
4	-
5	-

 

4

-

5

F

 

3

T

2

T

1

N7

2

1	NULL
2	-
3	-
4	-
5	-

 

3

1	NULL
2	-
3	-
4	-
5	-

 

4

-

5

T

 

3

F

1	N4
...	..

 

4

-

5

T

 

4

-

5

T

 

3

T

1	N1
..	..

 

4

T  终极答案

5

--

 

 

 

完整源码：

class Solution {

    private TreeNode ans;

    public Solution() {
        this.ans = null;
    }

    private boolean dfs(TreeNode root, TreeNode p, TreeNode q) {
        System.out.println("节点：" + root);
        if (root == null) {
            System.out.println("=====：null==>" + false);
            return false;
        }
        boolean lson = dfs(root.left, p, q);
        boolean rson = dfs(root.right, p, q);
        if ((lson && rson) || ((root.val == p.val || root.val == q.val) && (lson || rson))) {
            ans = root;
        }
        boolean b=lson || rson || (root.val == p.val || root.val == q.val);
        System.out.println("=====："+root+ "==>" + b);
        return b;
    }

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        this.dfs(root, p, q);
        return this.ans;
    }

    public static void main(String[] args) {
        TreeNode n1 = new TreeNode(3, 5, 1);
        n1.left.left = new TreeNode(6);
        n1.left.right = new TreeNode(2, 7, 4);
        n1.right = new TreeNode(1, 0, 8);

        TreeNode n = new Solution().lowestCommonAncestor(n1, new TreeNode(8), new TreeNode(7));
        System.out.println(n);
    }

}

class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;

    public TreeNode(int val) {
        this.val = val;
    }

    public TreeNode(int val, int left, int right) {
        TreeNode l = new TreeNode(left);
        TreeNode r = new TreeNode(right);
        this.val = val;
        this.left = l;
        this.right = r;
    }

    @Override
    public String toString() {
        return "" + this.val;
    }
}