LeetCode 236. Lowest Common Ancestor of a Binary Tree

LeetCode 236. Lowest Common Ancestor of a Binary Tree

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.

题目大意：

找两个节点的最近祖先

思路：

• 如果p在左子树里，q在右子树里（或者反过来，p在右子树，q在左子树），说明当前节点为pq的最小祖先。
• 反之，说明pq要么都在左子树里，要么都在右子树里。需要分别遍历左子树和右子树。
• 用后续遍历的思想，从下往上，如果遇到了p（或q），就把p（或q）的值向上返回。以下图为例，p=3，q=9。现在要查找3和9的最近祖先。把3和9一层一层向上返回，直到6。此时6的左子树接收返回值3，右子树接收返回值9，说明6是3和9的最近祖先。

感觉不太好理解，或许看看代码能加深一下理解。另外递归确实不好写，java代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */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;        // 替换成下面的语句也可以        if((left == p && right == q)||(left == q && right == p)) {            return root;        }        return (left != null) ? left : right;    }}

注：学渣心里苦，不要学楼主，平时不努力，考试二百五，哭~