236. Lowest Common Ancestor of a Binary Tree（最低公共祖先，难理解）

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 5and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

 

• 树中 find val 三种写法
  
  • 前序
    
  • 前序
    
  • 后序：遍历二叉树中所有节点
    
 

在find函数的后序位置，如果发现left和right都非空，就说明当前节点是LCA节点，即解决了第一种情况：

在find函数的前序位置，如果找到一个值为val1或val2的节点则直接返回，恰好解决了第二种情况：

因为题目说了p和q一定存在于二叉树中(这点很重要），所以即便我们遇到q就直接返回，根本没遍历到p，也依然可以断定p在q底下，q就是LCA节点。

 

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (root == NULL) return NULL;
        // 前序遍历位置，一条绳上，找到 p\q 一定就是公共祖先，都不需要往下遍历。
        if (root == p || root == q) {
            return root;
        }
        // 如果 pq 在树的左右两边。
        TreeNode*  left = lowestCommonAncestor(root->left,p,q);
        TreeNode*  right = lowestCommonAncestor(root->right,p,q);
        // 都不为空，说明找到了祖先，否则继续去左右子树寻找
        if (left != NULL && right != NULL) {
            return root;
        } else if (left != NULL) { 
            return left;
        } else if (right != NULL) {
            return right;
        }
        return NULL;
    }
};