Lowest Common Ancestor of a Binary Tree Leetcode

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/aprilyang/p/6357308.html

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#数据结构与算法

本文介绍了一种使用递归方法寻找二叉树中两个节点的最低公共祖先的有效算法，并给出了具体的实现代码。

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 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.

这道题我应该是做过的。。。然而一开始还是没做出来。。。究其原因，就是一开始就用了stack而没有往递归的方向想，用递归是最简单的。

最近树的题目真的是生疏了，要多加练习才好。

这里有个需要注意的地方就是，比较的时候一定要TreeNode直接比较，不要比较val，因为给的就是treenode，要treenode相等才是相等，不是值相等就是相等，因为还会有值一样的情况，这样就很难办。

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if (root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) {
            return root;
        }
        if (left == null && right != null) {
            return right;
        }
        if (left != null && right == null) {
            return left;
        }
        return null;
    }
}

但这个代码写的其实蛮冗余的，可以这样写

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) {
            return root;
        }
        return left == null ? right : left;
    }
}