leetcode 235&236

235

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

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).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

该题解法应该依赖于二叉搜索树的性质。
注意：是求p和q的最近的共同祖先（可以是p和q本身），并不是p和q值最小的祖先

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode small = p.val <= q.val?p:q;
        TreeNode large = p.val > q.val?p:q;

        //one left, one right 
        if(small.val <= root.val && large.val>= root.val ) return root;
        //both in right subtree
        if(small.val >= root.val) return lowestCommonAncestor(root.right, p, q);
        //both in left 
        return lowestCommonAncestor(root.left, p,q);

    }
}

以上判断过程可以继续优化
取0和1正1负，都返回root
都大于1的话，说明都在同一边的子树上

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    return (root.val - p.val) * (root.val - q.val) < 1 ? root :
           lowestCommonAncestor(p.val < root.val ? root.left : root.right, p, q);
}

如果题目改成求p和q值最小的共同祖先，解法类似

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        TreeNode small = p.val <= q.val?p:q;
        TreeNode large = p.val > q.val?p:q;

        //one left, one right 
        if(small.val <= root.val && large.val>= root.val ) return root;
        //both in right subtree
        if(small.val >= root.val) return root;
        //both in left 
        return lowestCommonAncestor(root.left, p,q);

    }

236

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.

这里的树是任意的二叉树
我的递归解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null) return null;
        if(root==p || root==q){
            // TreeNode left = lowestCommonAncestor(root.left, p, q);
            // TreeNode right = lowestCommonAncestor(root.right, p,q);
            return root;
        }else{
            TreeNode left = lowestCommonAncestor(root.left, p, q);
            TreeNode right = lowestCommonAncestor(root.right, p,q);
            if(left!=null&&right!=null) return root;
            if(left!=null) return left;
            if(right!=null) return right;
            return null;
        }
    }
}

这里其实有一个假设，就是假设p和q是合法的，也就说说p和q都在树上。
思路是，一个节点为根的子树上存在p或q，返回存在的点，否则返回null。最终到根上，必定返回的是最下的共同祖先。
如果该节点是p或q中的一个，直接返回根。这里如果子树种还有另外一个，根就是共同的祖先，直接返回没毛病。如果子树种没有了另一个，返回根就是返回了存在的。