236. Lowest Common Ancestor of a Binary Tree

Posted 2020-09-28 白天黑夜每日c

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.

二刷 ４０％

思考：遍历到节点ｎｏｄｅ，要判断它是否是ｐ、q的ＬＣＡ，想知道ｎｏｄｅ是否等于p、q，以及ｎｏｄｅ的左子树和右子树里包含p,q的情况。

建一个helper function，返回以ｎｏｄｅ为根的树里包含p、q的情况。

     若返回１，表示包含p; 若返回２，表示包含ｑ; 若返回３，表示包含p和ｑ; 若返回０，表示都不包含。

遍历到ｎｏｄｅ，可以分这几种情况：

     如果 node == p,再看ｎｏｄｅ的左右子树里是否包含q，如果包含返回３，并且把ｎｏｄｅ设为ｒｅｓｕｌｔ

     如果 node == q,再看ｎｏｄｅ的左右子树里是否包含p，如果包含返回３，并把ｎｏｄｅ设为ｒｅｓｕｌｔ

     如果 node != p 且node != q

         看ｎｏｄｅ的左子树，如果返回３(包含p、q)，则返回３（但不需要设置result, 因为ｒｅｓｕｌｔ应该在ｒｏｏｔ的左子树里）

         同理看ｎｏｄｅ的右子树，如果返回３，则ｒｏｏｔ也返回３

         如果helper(root.left) + helper(root.right) == 3，说明p和q一个在ｒｏｏｔ的左子树里，一个在右子树里，把ｒｏｏｔ设为ｒｅｓｕｌｔ，返回３

     所有为３的情况都返回了，不为３的话返回helper(root.left) + helper(root.right)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode node = null;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        findNode(root, p, q);
        return node;
    }
    public int findNode(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return 0;
        }
        int re = 0;
        if (root == p) {
            re = 1 + findNode(root.left, p, q) + findNode(root.right, p, q);
        } else if (root == q) {
            re = 2 + findNode(root.left, p, q) + findNode(root.right, p, q);
        } else {
            int left = findNode(root.left, p, q);
            if (left == 3) {
                return 3;
            }
            int right = findNode(root.right, p, q);
            if (right == 3) {
                return 3;
            }
            re = left + right;    
        }
        if (re == 3) {
            node = root;
        }
        return re;
    }
}

 recursively root,检查root是否是pq，root的左子树是否有pq，root的右子树是否有pq。 18%

如果左子树不含pq，lowestCommonAncestor(root.right,p,q)返回空，LCA在右子树中。

递归，检查root，检查root.left, 检查root.right

**这种办法是基于p、q一定在ｒｏｏｔ中

 

/**
 * 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) return root;
        TreeNode left=lowestCommonAncestor(root.left, p, q), right = lowestCommonAncestor(root.right, p, q);
        if(left!=null && right!=null) return root;
        if(right==null) return left;
        if(left==null) return right;
        return null;
    }
}