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.

找两棵树的最低父节点

自己做出来了一个递归的解法 但是stackoverflow了 testcase的树深度特别大 下面的解法会占用大量内存 但是可以通过testcase 

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {    Map<TreeNode, TreeNode> parent = new HashMap<>();    Deque<TreeNode> stack = new ArrayDeque<>();    parent.put(root, null);    stack.push(root);    while (!parent.containsKey(p) || !parent.containsKey(q)) {        TreeNode node = stack.pop();        if (node.left != null) {            parent.put(node.left, node);            stack.push(node.left);        }        if (node.right != null) {            parent.put(node.right, node);            stack.push(node.right);        }    }    Set<TreeNode> ancestors = new HashSet<>();    while (p != null) {        ancestors.add(p);        p = parent.get(p);    }    while (!ancestors.contains(q))        q = parent.get(q);    return q;}

遍历整棵树把对应关系存入parent key是节点 value是父节点 

然后获取p的所有父节点 对q 从下到上逐层获取父节点 找到第一个与p的公共父节点 也就是最低的公共父节点 

下面是递归的解法

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {    if(root == null || root == p || root == q)  return root;//1.    TreeNode left = lowestCommonAncestor(root.left, p, q);    TreeNode right = lowestCommonAncestor(root.right, p, q);    if(left != null && right != null)   return root;//2.    return left != null ? left : right;//3.}

效率比较高 但是之前觉得不是很好理解

四个月之后再看 感觉也不难理解

1.如果p是根节点 那么直接返回 因为q肯定是p的子节点

2.如果在当前节点下找到了p和q 那当前节点就是最低父节点 返回即可

3.如果在当前节点下 没找到q 那说明q是p的子节点 返回p即可