236. Lowest Common Ancestor of a Binary Tree

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.

解法1：自己一开始用的办法，比较慢

/**
 * 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 {
    bool hasnode(TreeNode* root,TreeNode* p)
    {
        if(!root) return false;
        if(root==p)
        return true;
        return hasnode(root->left,p)||hasnode(root->right,p);
    }
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root) return NULL;
        if(root==p||root==q) return root;
        if((hasnode(root->left,p)&&hasnode(root->right,q))||(hasnode(root->left,q)&&hasnode(root->right,p)))
        return root;
        TreeNode* r=lowestCommonAncestor(root->left,p,q);
        return r==NULL?lowestCommonAncestor(root->right,p,q):r;
    }
};

转载于 点击打开链接

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root||root==p||root==q) return root;
        TreeNode* left=lowestCommonAncestor(root->left,p,q);
        TreeNode* right=lowestCommonAncestor(root->right,p,q);
       // if(left&&right) return root;
        //if(left) return left;
        //else return right;以上等价于下面的return语句
        return !left?right:!right?left:root;
    }

如果当前节点包含p，而它同一层的另一子树包含q,返回两者的父节点。否则，只在某一子树下接着遍历。