leetcode236

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

        _______3______

       /              

    ___5__          ___1__

   /              /      

   6      _2       0       8

         /  

         7   4

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1

Output: 3

Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4

Output: 5

Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself

             according to the LCA definition.

Note:

* All of the nodes' values will be unique.

* p and q are different and both values will exist in the binary tree.

 

 

postOrder分治法。

1.返回TreeNode，逻辑抽象。是P传P(root)，是Q传Q(root)，左右都有东西传说明这个是lca传(root)，左右只有一个有东西说明这个是更高级的ca传有东西的那个结果，左右都没东西说明这个什么都不是传null(左右随便选一个传) 

 

2.返回类型较复杂，逻辑清楚。hasP, hasQ, lca。逻辑是hasP hasQ根据左右还有自己的状态来得到，lca是在本层hasP&&hasQ的时候更新，下面有就帮忙传上去，下面没有就传自己。

 

 

法1：

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

        if (root == p || root == q) {
            boolean rootright = true;
            return root;
        }
        if (left != null && right != null) {
            boolean notempty = true;
            return root;
        }

        return right == null ? left : right;
    }
}

法2：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private class ResultType{
        boolean hasP;
        boolean hasQ;
        TreeNode lca;
        public ResultType(boolean hasP, boolean hasQ, TreeNode lca) {
            this.hasP = hasP;
            this.hasQ = hasQ;
            this.lca = lca;
        }
    }
    
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        ResultType rt = helper(root, p, q);
        return rt.lca;
    }
    
    private ResultType helper(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return new ResultType(false, false, null);
        }
        // if (root == p) {
        //     return new ResultType(true, false, null);
        // }
        // if (root == q) {
        //     return new ResultType(false, true, null);
        // }
        // P1: 一个是不要处理的这么细，拿root和pq是否相等也这样处理，尽可能包含在下面的递归里。另一个是hasP本来就要考虑当前root是不是的情况，如果左右都没有p但本root就是p那也得是hasP啊。这样才能正确处理到p是q的祖先的情况。
        ResultType left = helper(root.left, p, q);
        ResultType right = helper(root.right, p, q);
        boolean hasP = left.hasP || right.hasP || root == p;
        boolean hasQ = left.hasQ || right.hasQ || root == q;
        TreeNode lca = null;
        if (hasP && hasQ) {
            if (left.lca != null) {
                lca = left.lca;
            } else if (right.lca != null) {
                lca = right.lca;
            } else {
                lca = root;
            }
        }
        return new ResultType(hasP, hasQ, lca);
    }
}